Address from the President of the IChO-2007 ...... 2 Overview of the IChO-2007 ...... 4 Participating Countries ...... 6 Programs ...... 7 Minutes of the Business Sessions of the 39th IChO in Moscow ...... 10 Practical Examination Problems ...... 17 Theoretical Examination Problems ...... 30 Answer Sheets, Solutions, and Grading Schemes ...... 51 Statistical Analysis of the Problems ...... 112 Results per Student ...... 117 Detailed Results per Country ...... 122 Comparative Analysis of the IChO Results from 1990 to 2007 ...... 132 List of Mentors, Observers, and Guests ...... 133 IChO-2007 Budget ...... 141 Participation fees for the IChO-2007 ...... 142 Preparations ...... 144 People who made the IChO-2007 possible ...... 145

1 Address from the President of the IChO-2007

Dear friends!

The 39th International Chemistry Olympiad is in the past. What will you remember from July 2007? I really hope your memory recalls bright moments of the Opening and Closing Ceremonies, nice sites of Moscow, unforget- table time spent with many old and new friends. And of course, mentors will remember intriguing discussions of examination tasks, whereas students – huge efforts to solve them. I am really sure everybody agrees that the 39th IChO became an important step in the development of International Chemistry Olympiads and science education as a whole.

Many organizations and individuals significantly contributed to the 39th IChO suc- cess. I have a pleasure to thank the Ministry of Education and Research of Russian Federation and personally the Minister A. Fursenko for serious attention to the Olympiad, which we felt at every stage of preparation.

I address warmest words of gratitude to the Rector of Moscow State University, Full Member of Russian Academy of Sciences, Professor V. Sadovnichy. Organization of IChO required coordinated efforts of various divisions and services of the Moscow University, and we always got positive and benevolent response and readiness to help.

MSU Chemistry Department was the host organization of the 39th IChO, and thus delegated the maximal number of members to the Science Committee, as well as to the working groups which assisted during examinations and other Olympiad events. I thank everyone who contributed to the IChO. Without your enormous efforts, dedica- tive work and high professionalism, this Olympiad would never have come true.

We are extremely grateful to all the sponsors. Their donations allowed conducting the Olympiad at a high technical level. I am absolutely sure that it was a very useful

2 investment in the growth of talented young people from all over the world, and the companies will gain numerous benefits in future.

It is my pleasure to thank all the students for excellent participation in IChO and following the standards of fair competition. My congratulations on excellent results achieved at quite complicated sets of practical and theoretical problems. And of course I am very grateful to all mentors, scientific observers, guests and guides. Your positive attitude, creativity and enthusiasm produced warm and friendly atmos- phere which we all experienced throughout the Olympiad.

I really hope that all of you enjoyed the time spent in Moscow, and at MSU Chemis- try Department, in particular. As the Dean of Chemistry Department and President of the 39th IChO I have a pleasure to say to all the participants: You are very much welcome back!

With best wishes

Valeriy V. Lunin Professor Dean of MSU Chemistry Department Full Member of Russian Academy of Science President of the 39th IChO

3 OVERVIEW of the ICHO-2007

Period July 15-24, 2007

Venue Chemistry Department, M.V. Lomonosov Moscow State University

Official Airports Sheremetievo International Domodedovo International

Students Location Hotel ‘Olympiets’, 5 km away from Moscow

Mentors, Scientific Observers and Guests Location Hotel “Holiday Inn Sokolniki”

Host Organization Chemistry Department, M.V. Lomonosov Moscow State University

Supported by Government of Russian Federation Ministry of Education and Science, RF M.V. Lomonosov Moscow Sate University

Main Sponsors “Basic Element” Company, ”Lukoil”-neftekhim Vladimir Potanin Foundation

Sponsors Norilsk Nickel; DuPont; BASF; Schlumberger; ESN Group of Companies; Procter and Gamble; JSC “Sibur”; Eurocement Group; Tokyo Boeki Ltd., Renova Group of Companies; Helios IT-Operator; Analytik Jena; Institute for New Carbon Materials; Russian Chemist Union; Bruker Companies; Oil Transporting JSC “Transneft”; Publishing House “Drofa”; Bayer MaterialScience; BP Traiding Ltd.; Dr. Barbara Mez-Starck Foundation; Lhoist Group; ACD/Labs; BioChemMac Group of Companies 4 Information Sponsors Publishing holding “Career-Press”; “Chemistry and Life”; The Chemical Journal; “Nanometer” portal

Participation Countries 67 participating, 1 observing (Nigeria), new participant – Moldova Students 256 Mentors 132 Scientific Observers 44 Guests 15

Results Gold Medals 31 Silver Medals 56 Bronze Medals 71 Honorable Mentions 10

Operational Stuff 230 Organizing Committee 20 Science Committee 34 Secretariat 8 Technical Committee 19 Lab Instructors 24 Lab Assistants 21 Spectrophotometer Operators 26 “Catalyzer” team 15 Guides 72

5 PARTICIPATING COUNTRIES

Country Code Country Code

Argentina ARG Kyrgyzstan KGZ Armenia ARM Latvia LVA Australia AUS Lithuania LTU Austria AUT Malaysia MYS Azerbaijan AZE Mexico MEX Belarus BLR Moldova MDA Belgium BEL Mongolia MNG Brazil BRA Netherlands NLD Bulgaria BGR New Zealand NZL Canada CAN Norway NOR China CHN Pakistan PAK Chinese Taipei TPE Peru PER Croatia HRV Poland POL Cuba CUB Portugal PRT Cyprus CYP Romania ROU Czech Republic CZE Russian Federation RUS Denmark DNK Saudi Arabia SAU Estonia EST Singapore SGP Finland FIN Slovakia SVK France FRA Slovenia SVN Germany DEU Spain ESP Greece GRC Sweden SWE Hungary HUN Switzerland CHE Iceland ISL Tajikistan TJK India IND Thailand THA Indonesia IDN Turkey TUR Iran IRN Turkmenistan TKM Ireland IRL Ukraine UKR Israel ISR United Kingdom GBR Italy ITA United States USA Japan JPN Uruguay URY Kazakhstan KAZ Venezuela VEN Korea Republic KOR Vietnam VNM Kuwait KWT

Nigeria – observing country (first year)

6 PROGRAMS Students and guides

Arrivals, Registration at Chemistry Department, MSU July 15 Whole day Transfer to Olympiets Hotel Sunday 18.00-19.00 Welcome Dinner, Olympiets

9.00 Buses depart to the Opening Ceremony, MSU 11.00-13.00 Opening Ceremony, MSU Intellectual Centre – Fundamental Library July 16 13.00-15.00 Welcome Reception, MSU Main Building Monday 15.00-18.00 Moscow City tour 18.00 Transfer to Olympiets 20.00-24.00 Disco

9.00 Buses depart to the North Shipping Terminal July 17 10.00-17.00 Boat trip along Moscow-river, lunch included Tuesday 17.00 Transfer to Olympiets 19.00-21.00 Lab safety instructions in groups

8.00 Buses depart to the Practical Exam, MSU July 18 10.00-15.00 Practical Exam, MSU Chemistry Department Wednesday 17.00-18.30 Walk-around MSU Campus & Lenin Hills 19.00-22.00 Circus show

July 19 8.30-19.30 Whole-day excursion to Sergiev Posad Thursday

8.00 Buses depart to the Theoretical Exam, MSU 10.00-15.00 Theoretical Exam, MSU Chemistry Department July 20 Friday 17.00-18.00 Walk to "Vorobjovi Gori" berth 18.00-22.00 Re-union Party on a boat, dinner included 22.00 Transfer to Olympiets

8.40 Buses depart to excursions July 21 10.00-14.00 Excursion to Kremlin Saturday 15.00-18.00 Excursion to Moscow Zoo 18.00 Transfer to Olympiets

9.00-13.00 Paintball & "Adventurer" game July 22 14.00-18.00 Free time, sports activities Sunday 19.00-23.00 Disco

9.00-12.00 Free time 12.40 Buses depart to the Closing Ceremony, MSU 15.00-18.00 Closing Ceremony, MSU Assembly Hall July 23 Monday 18.00-18.30 Walk-around MSU Campus 18.30 Buses depart to the Farewell Banquet 19.00-23.00 Farewell Banquet, Russian Academy of Sciences 23.00 Transfer to Olympiets

9.00-12.00 Check-out July 24 Whole day Departures Tuesday 12.00 Late departures – staying in Olimpiets

7 Mentors and observers

Arrivals, Registration at Chemistry Department, MSU July 15 Whole day Transfer to Holiday Inn Sunday 22.00-24.00 Get-together party, Holiday Inn, Vorobjevy Gory Hall

9.00 Buses depart to the Opening Ceremony, MSU 11.00-13.00 Opening Ceremony, MSU Intellectual Centre – Fundamental Library 13.00-15.00 Welcome Reception, MSU Main Building July 16 15.00-16.00 Lab inspection Monday 16.30 Buses depart to Holiday Inn, distribution of Practical Exam 18.00-19.00 Consultation with tasks authors, Polyanka Hall 20.00-24.00 The 1st Jury Meeting, Sokolniki 2 Hall

July 17 Whole day Translation of Practical Exam, Holiday Inn, Sokolniki Hall Tuesday 22.00-24.00 Informal party, Holiday Inn, Vorobjevy Gory Hall

9.30-14.00 Excursions in groups: Kremlin/City tour July 18 17.30-19.00 Consultation with tasks authors, Arbat Hall Wednesday 20.00-24.00 The 2nd Jury Meeting, Okhotnyi Ryad & Vorobjevy Gory Halls

July 19 Whole day Translation of Theoretical Exam, Holiday Inn, Sokolniki Hall Thursday 22.00-24.00 Informal party, Holiday Inn, Vorobjevy Gory Hall

8.00-18.00 Whole-day excursion to Sergiev Posad, lunch included July 20 18.00-22.00 Re-union Party on a boat, dinner included Friday 22.00 Transfer to Holiday Inn

9.30-14.00 Excursions in groups: City tour/Kremlin July 21 15.00-19.00 Grading of students' works Saturday 20.00-22.00 The 3rd (Business) Jury Meeting, Sokolniki 1 Hall 22.00-24.00 Informal party, Holiday Inn, Vorobjevy Gory Hall

8.00-19.00 Arbitration in groups, Arbat Hall July 22 20.00-21.00 Meeting with Minister of Education, RF and sponsors, Sokolniki 1 Hall Sunday 21.00-22.00 The 4th Jury Meeting, Allocation of medals, Sokolniki 1 Hall 22.00-24.00 Informal party, Holiday Inn, Vorobjevy Gory Hall

10.00-12.30 Free time, souvenir hunting 13.30 Buses depart to the Closing Ceremony, MSU 15.00-18.00 Closing Ceremony, MSU Assembly Hall July 23 18.00-18.30 Walk-around MSU Campus Monday 18.30 Buses depart to the Farewell Banquet 19.00-23.00 Farewell Banquet, Russian Academy of Sciences 23.00 Buses depart to Holiday Inn 24.00-02.00 Informal Farewell Party, Holiday Inn, Vorobjevy Gory Hall

9.00-12.00 Check-out July 24 Whole day Departures Tuesday 12.00 Late departures – transfer to Olimpiets hotel

8 Guests

Arrivals, Registration at Chemistry Department, MSU July 15 Whole day Transfer to Holiday Inn Sunday 22.00-24.00 Get-together party, Holiday Inn, Vorobjevy Gory Hall

9.00 Bus departs to the Opening Ceremony, MSU July 16 11.00-13.00 Opening Ceremony, MSU Intellectual Centre – Fundamental Library Monday 13.00-15.00 Welcome Reception, MSU Main Building 15.00-18.00 Moscow City tour

10.00-14.00 Excursion to Kremlin, the Diamond Hall July 17 15.00-19.00 Free time, souvenir hunting Tuesday 22.00-24.00 Informal party, Holiday Inn, Vorobjevy Gory Hall

10.00-13.00 Excursions to the Tret’yakov gallery July 18 13.00-15.00 Lunch, "Godunov" restaurant, center of Moscow Wednesday 15.00 Transfer back to the Holiday Inn.

9.00-14.00 Excursion to the Kolomenskoe park July 19 17.30 Bus departs to the Moscow Circus, Tsvetnoi blvd. Thursday 19.00-22.00 Circus show 23.00-24.00 Informal party, Holiday Inn, Vorobjevy Gory Hall

8.00-18.00 Whole-day excursion to Sergiev Posad July 20 18.00-22.00 Re-union Party on a boat Friday 22.00 Transfer to Holiday Inn

10.00-14.00 Walk in the center of Moscow, Arbat July 21 15.00-19.00 Free time Saturday 22.00-24.00 Informal party, Holiday Inn, Vorobjevy Gory Hall

9.00-13.00 Excursion to the Cathedral of Christ the Savior July 22 Sunday 15.00-19.00 Excursion to the Izmailovo exhibition 20.00-21.00 Meeting with the Minister of Education, RF and sponsors, Sokolniki 1 Hall

9.00-12.30 Free time, souvenir hunting 13.30 Buses depart to the Closing Ceremony, MSU 15.00-18.00 Closing Ceremony, MSU Assembly Hall July 23 18.00-18.30 Walk-around MSU Campus Monday 18.30 Buses depart to the Farewell Banquet 19.00-23.00 Farewell Banquet, Russian Academy of Sciences 23.00 Buses depart to Holiday Inn 24.00-02.00 Informal Farewell Party, Holiday Inn, Vorobjevy Gory Hall

9.00-12.00 Check-out July 24 Tuesday Whole day Departures 12.00 Late departures – transfer to Olimpiets hotel

9 Minutes of the Business Sessions of the 39th IChO in Moscow

MINUTES of the BUSINESS PART of the 1st JURY SESSION at the 39th ICHO/MOSCOW July 16th, 2007 (20.30 – 21.50) Chair of the BUSINESS PART: Manfred Kerschbaumer

1. Presentation of the agenda for the meeting. 2. Number of delegations present at 20.30: 60 3. Short report of the ISC Meeting in December 2006: Detailed minutes of the ISC-meeting may be found in the appendix of the Pre- paratory Problems for Moscow 2007; ISC is aware of the situation with level 3 tasks – work on that has already be- gun. In the business session of IChO (Jury session #3) Andras Kotschy (Hungary) will distribute a proposal by Gabor Magyarfalvi (Hungary) for a new syllabus. 4. Change of regulations: possibility of a 2nd Scientific Observer Regulations § 3, 1: Each participating country's delegation may consist of four competitors and two accompanying persons (also known as mentors). Countries may include one scientific observer in their delegation. Proposal: Each participating country's delegation may consist of four competitors and two accompanying persons (also known as mentors). Countries may include two scientific observer in their delegation. Comments: Anton Sirota (Slovakia): strongly against it, regulations should not be changed within such a short period, problems with financ- ing, no necessity for a 2nd scientific observer if task are shorter; I-Jy Chang (Taiwan): 2nd scientific observer extremely important for Asian countries due to time consuming translation work; Carlos Castro-Acuna (Mexico): 2nd scientific observer decreases injustice in connection with language groups; Marek Orlik (Poland): against it, because too many people in the Jury sessions;

10 Azerbaijan: supports proposal, because in case of changing the mentors (“generation change”) 2nd scientific observer may learn a lot; Voting: A qualified majority would be 45 votes out from possible 66 (countries)

Result: 50 votes supporting the proposal ⇒ CHANGE ACCEPTED 5. Change of regulations: another definition of Honorable Mentions Regulations § 15, 5: An honorable mention is awarded to competitors who do not receive a medal, but gain full marks for at least one problem. Proposal: An honorable mention is awarded to competitors who are among the best 10% of non medalists. Comments: Carlos Castro-Acuna (Mexico): his proposal, more justice for students who miss the bronze medals very tightly Several other comments in favour of the change. Voting: A qualified majority would be 45 votes out from possible 66 (countries)

Result: 47 votes supporting the proposal ⇒ CHANGE ACCEPTED 6. Agenda for the Business Session of the IChO (draft) Ø Information by the Minister of Education and Science of Russia Ø Election of new SC-members: one from Europe, two from Pacific Rim (proposals!) Ø Distribution of “Gabor’s” proposal concerning new syllabus and comments on it Ø IUPAC-support Ø Future Olympiads (presentation from Hungary) Ø Any other business

11 MINUTES of the BUSINESS SESSION of the 39th ICHO/MOSCOW July 21st, 2007 (20.15 – 22.15) Chair of the BUSINESS SESSION: Manfred Kerschbaumer

63 delegations are present.

Dr. Kerschbaumer presents the agenda of the meeting: · Election of new SC-members: one from Europe, one from Asia, one from Pacific Rim · Level and length of competition – new syllabus: Information about “Gabor’s” proposal · Level and length of competition – proposal to cut down the number of characters · IUPAC-support · Future Olympiads · Information about arbitration

· Election of new SC-members: The following proposals are presented to the International Jury: EUROPE: Wesley Brown (Ireland) Kurt Nielsen (Denmark) ASIA: Duckwhan Lee (Korea) PACIFIC RIM: Mark Ellison (Australia) As Duckwhan Lee and Mark Ellison are the only candidates from their respective regions, no election is necessary, they will be members of the SC for the next two years. After short self presentations from the two European candidates, the jury elects Wesley Browne with 36 votes (Kurt Nielsen: 19 votes) as the third member of the SC from Europe. Dr. Kerschbaumer thanks the parting member of SC, I-Yi Chang from Taiwan, Geoff Salem from Australia, and Paraic James from Ireland. The members of the current SC are listed in appendix A.

12 · Level and length of competition – new syllabus: Manfred Kerschbaumer reports from the SC-meeting in December 2006 where the SC stated that in all the years before the regulation concerning the allowed number of level 3 topics was violated. So it was the duty of the SC to try and change this abuse. Andras Kotschy from Hungary presents the proposal by Gabor Magyarfalvi which changes the old syllabus in its basics. Instead of topics belonging to three different levels, the new syllabus contains a section “concepts and skills” and a section “factual knowledge”. In both sections there is a list of concepts and skills or factual knowledge respectively which can be expected from the participants to be familiar with, and a list which must be covered in the preparatory problems. A hard copy with the details of this proposal is distributed. The members of the Inter- national Jury are requested to give comments.

· Level and length of competition - proposal to cut down the number of characters Manfred Kerschbaumer proposes to cut down the number of characters allowed in the theoretical and practical competition (25.000 each, § 13, 4 of the regulations) to 22.000. He also points out that a lower number of characters does not imply an eas- ier set of tasks because a guiding explanation needing more words may be simpler for a student than a short task with even shorter questions. The head mentor from Belgium (Sebastien Delfosse) proposes to decrease this number to 20.000 characters. The jury decides to vote about both proposals with 38 votes in favour. As a change of the allowed number of characters means a change in the regulations, a qualified majority (45 votes) is necessary. After some discussion voting was carried out: In favour to cut down to 20.000 characters: 27 votes In favour to cut down to 22.000 characters: 35 votes As this is not a qualified majority the current regulation (25.000) will not be changed.

· IUPAC-support Manfred Kerschbaumer reports about an offer from the president of IUPAC (Bryan O. Henry) in Korea to support the ICHO in some way. From December 2006 till June

13 2007, Dr. Kerschbaumer managed to negotiate an annual donation of US $ 10.000 to help countries with financial troubles. A “Memorandum of Understanding” was es- tablished (appendix B) which shows the details of a possible support for the countries mentioned above.

· Future Olympiads (presentation from Hungary) Andras Kotschy shows a detailed presentation about the 40th ICHO in Budapest. It will be held at the Eötvös Loránd University from July 12th to July 21st. Andras dis- tributes the issue No. 1 of the Catalyzer. Yoshiyuki Sugahara presents some details about the 42nd ICHO in Tokio in 2010. The then following host will be Turkey (2011) and USA (2012). John Kotz (USA) in- forms about the possible venue of the 44th ICHO: Boston or Chicago.

· Information about arbitration Sasha Gladilin refers about the procedure of arbitration the next day. This time the sequence is not alphabetically but is decided by drawing lots.

14 APPENDIX A:

STEERING COMMITTEE OF THE ICHO as constituted on July 22, 2007 Elected members:

Representatives of Europe: Manfred Kerschbaumer (Austria), chairman; e-mail: [email protected] Alexander K. Gladilin (Russia); e-mail: [email protected] Wesley Browne(Ireland); e-mail: [email protected]

Representative of the Americas Carlos Castro-Acuna (Mexico); e-mail: [email protected]

Representatives of Asia and the Pacific Rim: Marc Ellison (Australia); e-mail: [email protected] Duckwhan Lee (Korea); e-mail: [email protected]

Non-elected members:

Representatives of the organizers: Valerie Lunin (Russia); e-mail: [email protected] Andras Kotschy (Hungary); e-mail: [email protected] Peter Wothers (United Kingdom); e-mail: [email protected] Representative from Japan

Experts: Wolfgang Hampe (Germany); e-mail: [email protected] Gabor Magyarfalvi (Hungary); e-mail: [email protected] Kurt Nielsen (Denmark); e-mail: [email protected] Anton Sirota (Slovakia); e-mail: [email protected]

15 APPENDIX B: Memorandum of Understanding (IUPAC-support for the IChO)

Ø IUPAC will support the International Chemistry Olympiad by an annual donation of US$ 10 000. This grant can be terminated by IUPAC after giving notice 12 months in advance of the next IChO. Ø The president of IUPAC (or nominee) will be invited to the closing Ø ceremony of the IChO to give a short (5 to 10 minutes) address and to take part in the gold medal presentations. Ø The first financial support will be for the 40th IChO in Budapest (Hungary) 2008. Ø The money will be transferred to an account of the annual host (2008: Hun- gary). Ø Countries who wish to apply for support will make a written application (non- formal) to the chair of the Steering Committee (SC) of the IChO. The application must contain the amount of money needed, and a (detailed) description of the problems of raising money for the IChO in that particular country. Deadline for the application is November 30th. Ø The chair of the Steering Committee will present all applications to the SC in its annual meeting at the beginning of December the year preceding the IChO in question. Ø The SC will make decisions regarding the distribution of the money consider- ing the following items: The money will be distributed to • countries which cannot afford the participation fee and therefore cannot participate in the IChO • countries which can not travel with a team of four students, • countries which can not travel with two mentors. Ø If the US$ 10 000 support is not fully allocated for a particular IChO, the un- used part of the donation will be transferred to the next host. Ø The IChO will provide IUPAC with a summary report of the distributions made from the IUPAC grant.

Bryan R. Henry (Date) Manfred Kerschbaumer (Date) President IUPAC Chairman of the Steering Committee of the International Chemistry Olympiad

16 PRACTICAL EXAMINATION PROBLEMS

General Directions

· safety rules: follow them as in the Preparatory problems described, no eating or drinking is allowed in the lab. · violating safety rules: you get one warning, offend again: you are out. · problem booklet: 12 pages (incl. cover sheet and Periodic table of elements) with 2 problems. Start with problem 1. · time: 5 hours; 30 minutes warning before the end. · answer sheets: 5 pages (incl. cover sheet). · your name and student code: write it on every answer sheet. · answers: only in the appropriate places of the answer sheets, nothing else will be marked. Relevant calculations have to be shown. · use only the pen and calculator provided. · results: the number of significant figures in numerical answers must conform to the rules of evaluation of experimental error. Mistakes will result in penalty points even if your experimental technique is flawless. · burette: read it as accurately as possible. · more chemicals needed? Ask your lab assistant. No penalty for this. · Extra sample to be analyzed or broken column: a penalty of 10 marks. · questions concerning safety, apparatus, chemicals, organization, toilet break: ask your lab assistant. · chemical waste: put it only in the designated containers. · official English-language version available on request for clarification only. Ask your lab assistant. · after the stop signal put your answer sheets and spectra in the envelope (don’t seal), deliver them to your lab assistant. Keep the problem booklet together with the pen and calculator. · You must stop your work immediately after the stop signal has been given. A delay of 5 minutes will result in zero points for the current task. · During the Practical examination some of your glassware and plastics may have to be used more than once. Clean it carefully.

17 List of Chemicals

Reagent Quantity Placed in Labeled Task 1 Eluent 1 100 mL Amber glass bottle* Eluent 1 Eluent 1 1 mL Plastic microtube Eluent 1 Eluent 2 50 mL Amber glass bottle* Eluent 2 Eluent 2 1 mL Plastic microtube Eluent 2 Eluent 3 50 mL Amber glass bottle* Eluent 3 Eluent 3 1 mL Plastic microtube Eluent 3 0.5 М Carbonate buffer solution, pH 9.5 10 mL Glass vial NaHCO3 0.5 М Tris-HCl buffer solution, pH 8.5 10 mL Glass vial Tris-HCl Mixture of amino acids to be analyzed** 1.2 mL Plastic microtube A number between 301 and 600 Ellmann reagent: 0.2 М Phosphate buffer solu- 10 mL Glass vial DTNB tion containing 10 mM EDTA and 3 mM 5,5’- Dithiobis(2-nitrobenzoic acid), pH 7.0 Pauli’s reagent: solution of sodium 4-diazonium- 1 ml Plastic microtube Pauli benzenesulfonate in 0.1 M aqueous HCl Sodium hydroxide, 10% aqueous solution 10 mL Glass vial NaOH 10% 8-Hydroxyquinoline, 5.2 mM solution in etha- 5 ml Glass vial 8-HQ nol/n-butanol (9:1) mixture Sodium hypobromite, 0.24 M solution in 10% 1.2 ml Plastic microtube NaBrO aqueous NaOH 2,4,6–Trinitrobenzenesulfonic acid, 3.4 mM 1 mL Plastic microtube TNBS aqueous solution 8 M Aqueous urea solution 1 mL Plastic microtube Urea Task 2 HCl, standard solution, ~1 M (see exact value on 40 mL Amber glass vial HCl the label) NaOH (to be standardized) 200 mL Amber glass vial NaOH Powdery sample to be analyzed** 0.5 – 1 g 150 mL beaker covered H2O distilled 400 mL Plastic wash bottle H2O H2O distilled (shared between 2 students) 30 mL Glass drop bottle H2O H2O distilled (for common use) 5 L Bottle with tubing and H2O clamp on top of the bench NaH2PO4, 15% solution (shared between 2 stu- 20 mL Glass drop bottle NaH2PO4 15% dents) Bromocresol Green, 0.5% solution in 20% etha- 30 mL Glass drop bottle Bromcresol nol (shared among 3-4 students in a raw) green Thymolphthalein, 0.5% solution in ethanol 30 mL Glass drop bottle Thymolphtalein (shared among 3-4 students in a raw) K2C2O4, 15% solution (shared between 2 stu- 50 mL Amber glass vial K2C2O4 15% dents) *Fixed on the top shelf (do not try to remove), with connected tubing and clamp **10 marks penalty for an extra portion of the sample

Components of Eluents 1 to 3 Eluent 1: 0.1 M aqueous sodium citrate, 50 mM sodium chloride, 40 mM thiodiglycol, 1 mM caprylic acid, 0.1% Brij-35; pH 4.9. Eluent 2: 0.2 M aqueous sodium phosphate, 0.1% Brij-35; pH 7.0. Eluent 3: 0.2 M aqueous sodium hydroxide.

18 Apparatus and Suppliers

Item Quantity Test tube rack 1 Laboratory stand 1 Chromatography column with ion-exchange resin 1 Laboratory stand with white covering 1 Double clamp for burette 1 Ring for funnel 1 25 mL Burette 1 100 mL flask labeled “Waste” 1 100 mL Volumetric flask 2 100 mL Erlenmeyer flask 2 Syringe with needle 1 Graduated test tubes for collecting fractions and preparing mixtures 50 96-well plate 1 Pipettor (micropipette) with fixed volume of 0.1 mL 1 Disposable tips (in blue plastic cup) 20 Spectrophotometric cuvettes labeled “A1”, “B1”, “A2”, “B2”, “A3”, “B3” in cuvette 6 holder 10 mL Graduated plastic pipettes 3 10 mL Glass pipette 1 Pipette filler 1 3-Way Bulb 1 Glass rod 1 Filter funnel 1 Small funnel 1 60 mL Amber glass vials for combined fractions (peaks) 3 10 mL Measuring cylinder labeled “K2C2O4 15%” (shared between 2 students) 1 10 mL Measuring cylinder (shared between 2 students) 1 50 mL Measuring cylinder 1 100 mL Measuring cylinder labeled “H2O” (shared among 3-4 students in a row) 1 Plastic plate with filters*** (shared among 3-4 students in a row) 3 filters per student Heating plate (for common use in a fume hood) 6 plates per hood Rubber protection tips (for common use a fume hood) 6 pairs per hood Spectrophotometer (shared by a group of students; see the number of the spectrophotometer to be used at your bench “SP____”) Marker 1 Ruler 1 White sheet of paper 1 ***If needed, ask your lab assistant for extra filters.

19 Safety regulations, S-phrases, R-phrases

Disodium hydrogen phosphate R:36/37/38 S:26-36 Ethylenediaminetetraacetic acid, disodium salt R:36/37/38 S:26-36/37/39 Tris-HCl R:36/37/38 S:26-36 Arginine R:36 S:26 Cysteine R:22 Histidine S:22-24/25 Hydrochloric acid R:34-37 S:26-36-45 Sodium 4-diazoniumbenzenesulfonate R:1-37/37 S:26-36 Sodium hydroxide R:34-35 S:26-36-37/39-45 8-Hydroxyquinoline R:22-36/37/38 S:26-36/37 Ethanol R:11 S:7-16 Butanol-1 R:10-22-37/38-41-67 S:7/9-13-26-37/39-46 Sodium hypobromite R31-34 S:26-36-45 5,5’-Dithiobis(2-nitrobenzoic acid) R:36/37/38 S:26-36 2,4,6-Trinitrobenzene sulfonic acid R: 1-22-36/38-43 S: 26-36/37 Sodium chloride R:36 S:26 Thiodiglycol R:36 S:26 Caprylic acid R:34 S:26-27-45-36/37/39 Brij-35 R:36/37/38 S:26-36 Sodium dihydrogen phosphate S:22-24/25 Sodium carbonate R:36 S:22-26 Calcium carbonate R:41-37/38 S:26-39 Bromocresol Green S:22-24/25 Thymolphthalein S:22-24/25 Potassium oxalate R:34 S:26-27-36/37/39

Risk Phrases Indication of Particular Risks R1: Explosive when dry 35: Causes severe burns 10: Flammable 36: Irritating to the eyes 22: Harmful if swallowed 37: Irritating to the respiratory system 31: Contact with acids liberates toxic gas 41: Risk of serious damage to eyes 34: Causes burns 43: May cause sensitization by skin contact 67: Vapors may cause drowsiness and dizziness Combination of Particular Risks R24/25: Toxic in contact with skin and if swallowed 36/37/38: Irritating to eyes, respiratory system and skin 36/38: Irritating to eyes and skin 37/38: Irritating to respiratory system and skin

Safety Phrases

Indication of Safety Precautions S13: Keep away from food, drink and animal feeding 39: Wear eye/face protection stuffs 45: In case of accident or if you feel unwell, seek medi- 22: Do not breathe dust cal advice immediately (show label where possible) 26: In case of contact with eyes, rinse immediately with 46: If swallowed, seek medical advice immediately and plenty of water and seek medical advice show this container or label 27: Take off immediately all contaminated clothing 36: Wear suitable protective clothing Combination of Safety Precautions 7/9: Keep container tightly closed and in a well- 24/25: Avoid contact with skin and eyes ventilated place 36/37/39: Wear suitable protective clothing, gloves and eye/face protection 37/39: Wear suitable gloves and eye/face protection

20 Problem 1. ION-EXCHANGE CHROMATOGRAPHY OF AMINO ACIDS 20 points Ion-exchange chromatography is an important analytical and preparative method, which allows fractioning of charged substances. Interaction of ionic groups of the substances with counterions attached to the resin is behind the method. In this task you will have to carry out separation of a given mixture of three amino acids followed by quantitative assay of individual amino acids eluted from the column by using specific chromogenic reactions. Since queues of students are possible at spectropho- tometers, we strongly suggest you starting the exam with Problem 1.

O O NH2 O N OH HS OH HN NH OH NH N 2 NH2 NH2 H His Cys Arg

Three amino acids (see the structures above) are present in the mixture. These are histidine, cysteine, and arginine. Cross-linked sulfonated polystyrene is used as a cation-exchange resin (see the picture of the system below). At the beginning of the experiment the column is equilibrated with Eluent 1 (pH 4.9).

Procedure Chromatography. Step 1 Apply the given solution of a mixture of amino acids to the column. First, open the stopcock to allow the solvent in the column draining into the Erlenmeyer flask labeled “Waste” so that the solvent is level with the top of packing material, still preventing the resin surface from drying off. Close the stopcock and carefully add the analyzed solution to the top of the column by using a syringe. Open the stopcock and let the sample soak inside the gel (drain the solvent into the “Waste” flask). Close the stopcock and add about 1 mL of Eluent 1 (corresponds to ~ 1 cm of liquid in the column) by carefully releasing the tubing clamp. Attach the top joint tightly, fixing the column with one hand and the adaptor with the other (be sure that the joint is fitted closely to the column). Replace the “Waste” flask at the stand with the test tubes in the rack. Release the tubing clamp and open the stopcock to let the eluent flow down through

21 the column. Proceed with elution. (Always open the stopcock to start elution and close the stopcock to stop it).

Collect the fractions in the test tubes up to the volume of 2.5 mL (as shown in the Picture). If needed, label them with marker. After collecting each 4 to 8 fractions stop elution and carry out qualitative analysis of the collected samples.

Qualitative analysis of samples Qualitative assay of amino acids is based on the reaction of their α-amino groups with sodium 2,4,6–trinitrobenzene sulfonate (TNBS):

NO2 O2N NH + O 2 Na HOOC NH - NaHSO HOOC + O S NO2 + 3 NO R O R 2 NO2 O2N The assay is carried out in the wells of a polystyrene plate, each well corresponding to a definite test tube. Before starting the assay, mix 1 mL of TNBS solution with 10 mL of carbonate buffer solution and place 0.1 mL of the resulting mixture into half of the plate wells (from A1 to H5). Then add 0.1 mL of the analyzed fraction into a well. Start with A1 well and continue with B1, C1, etc (move top to bottom and left to

22 right). If an amino acid is present in the analyzed fraction, intense yellow coloration will develop in the corresponding well within 3 min. Use the coloration in the first well as the reference. To reliably estimate the coloration, place the plate on the white sheet of paper. Note: all aliquots of 0.1 mL should be added by using the pipettor. We expect you to use one tip for all fractions of a single peak.

1.1a Draw the profile of coloration intensity (qualitatively) on the plate sketch in the Answer Sheet. Use the following symbols: (–) – no coloration, 1 – weak coloration, 2 – moderate coloration and 3 – intense coloration. Keep drawing the profile during the whole chromatography process.

Continue collecting fractions and analyzing them until you get at least two wells with coloration as in A1 well, which will indicate that the first amino acid has left the column completely (end of the first peak).

Chromatography. Step 2 As soon as you are finished with collecting the first peak, change to Eluent 2. To do so, close the stopcock, fix the tubing clamp (Important!), disconnect the tubing leading to the bottle with Eluent 1 and connect the tubing leading to the bottle with Eluent 2. Attach the top joint tightly.

1.1b Indicate when the eluents have been changed by drawing lines between the corresponding wells on the plate sketch.

Continue elution, collecting fractions and carrying out qualitative analysis of samples as described above.

Chromatography. Step 3 As soon as you are finished with collecting the second peak, change to Eluent 3 as described in Step 2. Continue chromatography until the third amino acid leaves the column completely.

Stop chromatography by closing the stopcock and fixing the clamp.

23 Based on the results of qualitative analysis, choose the fractions which contain the amino acids.

1.1c Write down in the Answer Sheet the labels of wells corresponding to the chosen fractions. 1.2 Combine the fractions from each peak and measure the volumes of combined fractions using a measuring cylinder. Report the volumes of combined fractions ex- cluding amounts used for the qualitative analysis. Write down the obtained results in the Answer Sheet.

Pour combined fractions in the amber glass vials labeled “Peak 1”, “Peak 2”, “Peak 3”. Prepare samples for quantitative spectrophotometric analysis as described below.

When finished with Practical exam, close the vials and leave them on the table. The combined fractions will be subsequently analyzed by lab staff.

Spectrophotometric analysis For each probe, you should submit two cuvettes to the operator. Prepare the probes as follows. Important! When storing, always put cuvettes in the cuvette holder! All cuvettes have 2 ribbed and 2 working vertical surfaces. While operating with cuvettes, do not touch working surfaces, otherwise you may get incorrect values of absorbance.

Assay 1 (peak 1). Cysteine concentration is determined by the Ellmann reaction:

NO2

- O O O - - O S O - + O NH3 + H N S + - OH - 3 - O S S + O S O SH -H O - 2 O O NO 2 O NO 2

NO2

24 Test tube A1 (Reference). Place 0.1 mL of Eluent 1 from plastic microtube into a test tube and add 2.9 mL of Ellmann reagent (DTNB). Test tube B1 (Sample). Place 0.1 mL of the analyzed solution into a test tube and add 2.9 mL of Ellmann reagent (DTNB). Mix the contents of the test tubes thoroughly and transfer each mixture to the corresponding cuvettes labeled A1 (for reference) and B1 (for sample).

Assay 2 (peak 2). Determination of histidine concentration is based on the ability of imidazole moiety to react with diazonium compounds (Pauli reaction).

Test tube A2 (Reference). Place 2.8 mL of Tris-HCl buffer solution into a test tube, add 0.1 mL of Eluent 2 from plastic microtube and 0.1 mL of Pauli reagent. Test tube B2 (Sample). Place 2.8 mL of Tris-HCl buffer solution into a test tube, add 0.1 mL of the analyzed solution and 0.1 mL of Pauli reagent.

Mix the contents of the test tubes thoroughly and transfer each mixture to the corresponding cuvettes labeled A2 (for reference) and B2 (for sample).

Assay 3 (peak 3). Determination of arginine concentration is based on the ability of guanidinium moiety to react with some phenols under alkaline and oxidative conditions (Sakaguchi reaction).

Test tube A3 (Reference). Place 0.1 mL of Eluent 3 into a test tube and add 1.5 mL of 10% NaOH solution, 1 mL of 8-hydroxyquinoline solution and 0.5 mL of sodium hypobromite solution. Test tube B3 (Sample). Place 0.1 mL of the analyzed solution into a test tube and add 1.5 mL of 10% NaOH solution, 1 mL 8-hydroxyquinoline solution and 0.5 mL of sodium hypobromite solution.

Shake the test tubes vigorously for 2 min (Important!) and observe formation of or- ange color. Add 0.2 mL of 8 M urea solution to each test tube, mix the contents and transfer about 3 mL of each mixture to the corresponding cuvettes labeled A3 (for reference) and B3 (for sample).

25 All mixtures should be analyzed by spectrophotometry not earlier than 10 min and not later than 2 h after preparation. Submit the set of 6 cuvettes to the spectrophotometer operator. In case of a queue at the spectrophotometer, ask the operator to put your student code on the list at the signboard. You will be invited by the operator in due time. Meanwhile, you can answer the theoretical question and start fulfilling Problem No 2.

In case your sample(s) have not been subjected to studies within the proper time interval (which is quite improbable), prepare the sample(s) afresh.

Get the print-offs with the spectra of your samples and check it. Sign the print-offs and get the operator’s signature.

1.3 Determine absorbance at the corresponding wavelengths and calculate the content (in mg) of each amino acid in the mixture you were given. The optical length is 1.0 cm. Complete the Answer Sheets taking into account that one mole of each amino acid gives one mole of the corresponding product.

Reference data: The values of extinction coefficients: Molar masses of the amino acids. Product of Ellmann reaction: 13600 M–1cm–1 at Cysteine 121 g/mol 410 nm Histidine 155 g/mol Product of Pauli reaction: 6400 M–1cm–1 at 470 Arginine 174 g/mol nm Product of Sakaguchi reaction: 7700 M–1cm–1 at 500 nm

1.4. Draw three resonance structures of the species responsible for mixture coloration as a result of Ellmann reaction.

26 Problem 2. DETERMINATION OF CARBONATE AND HYDROGEN PHOSPHATE IN AN ABRASIVE SAMPLE 20 points

Na2CO3, CaCO3 and Na2HPO4 are the main constituents of abrasive powders. In this task you will have to determine carbonate and hydrogen phosphate ions in an abrasive sample by two acid-base titrations.

First, the exactly known amount of hydrochloric acid (taken in an excess) is added to the sample. As a result, hydrogen phosphates are transformed into H3PO4, whereas carbonates into CO2 which is further removed by boiling. Calcium ions initially present in the sample pass into the solution. Because of possible interference in subsequent analysis, they are precipitated as CaC2O4 and filtered off prior to the titration.

Next, the phosphoric acid formed is subjected to two titrations with pre-standardized NaOH solution and two different indicators: Bromocresol Green (BCG) and Thymol- – phthalein (TP). First, H3PO4 (and excess of HCl) is titrated to H2PO4 ion, the endpoint lying in slightly acidic medium (pH of ~4.5). It corresponds to the color change 2– of BCG from yellow to blue. The second titration proceeds till HPO4 is formed. The endpoint of the second titration corresponds to the color change of TP from colorless to blue (moderately alkaline medium, pH of ~10).

2– The content of CO3 ions in the sample is calculated by finding the difference between: a) the amount of the titrant equivalent to the initial amount of HCl (taken for the sample dissolution) and b) the amount of the titrant corresponding to the second endpoint (TP). 2– The content of HPO4 is calculated by finding the difference between the amounts of the titrant consumed to achieve two endpoints (TP and BCG).

Procedure

Step 1. Dissolution of the sample and removal of CO2 To the sample of the abrasive powder in a beaker covered with watch glass add 10.00 mL (exactly, with a pipette! Carefully, not removing the glass and avoid- ing losses because of splashing!) of ca. 1 mol/L HCl (see the exact concentration

27 of the acid on the label). After the most intensive stage of gas evolution is completed, heat carefully the solution in the beaker (covered with watch glass) on a heating plate until the gas evolution stops. Then bring the solution to boiling and boil it carefully for 2-3 min.

Step 2. Precipitation of calcium Remove the beaker from the plate; wash the steam condensate from the watch glass

down to the beaker with distilled water. Add 1-2 mL of 15% K2C2O4 solution with measuring cylinder. Put the beaker aside until the most part of the precipitate is formed (usually takes 10 to 20 min). Spend this time for standardization of the titrant solution of NaOH (see the procedure hereunder).

Step 3. Standardization of NaOH solution Place with a pipette 10.00 mL of HCl solution into a 100 mL volumetric flask, make up to the mark with distilled water and mix. Fill the burette with NaOH solution. Transfer with a pipette 10.00 mL of the diluted HCl solution from the volumetric flask to an Erlenmeyer flask. Add 1-2 drops of Thymolphthalein solution and titrate with NaOH solution until blue coloration stable on swirling for 5-10 s appears.

Here and after. Repeat the titrations as necessary. It is desirable that the highest and the lowest titrant volume values differ not more than by 0.10 mL. Report all the final volume values with 0.01 mL accuracy.

2.1a Complete the table in the Answer Sheet. 2.1b Calculate the concentration of NaOH solution (in mol/L).

Step 4. Filtering off calcium oxalate

After the most part of CaC2O4 precipitates filter the precipitate off collecting the filtrate into a 100 mL volumetric flask. Slight turbidity in the filtrate is admissible, since small amounts of calcium oxalate do not interfere in the titration. Wash the filter with distilled water; make up the solution in the flask to the mark with distilled water and mix. Put the used filter into the waste basket.

Step 5. Sample titration against Bromocresol Green

28 Transfer with a pipette a 10.00 mL aliquot of the sample solution coming from the step 4 from the volumetric flask to an Erlenmeyer one, and add 3 drops of BCG solution. Prepare in another Erlenmeyer flask a reference solution by adding 3 drops of

15 % NaH2PO4 solution and 3 drops of BCG solution to 15-20 mL of distilled water. Titrate the sample solution with NaOH solution until the color coincides with that of the reference solution.

2.2 Complete the table in the Answer Sheet.

Step 6. Sample titration against thymolphthalein Transfer with a pipette a 10.00 mL aliquot of the sample solution coming from the step 4 from the volumetric flask to an Erlenmeyer one. Add 2 drops of TP solution and titrate with NaOH solution until blue coloration stable on mixing for 5-10 s appears.

2.3 Complete the table in the Answer Sheet.

Step 7. Calculations

2– 2.4 Calculate the mass of CO3 in the sample. 2– 2.5 Calculate the mass of HPO4 in the sample.

Step 8. Additional questions to the problem Answer the additional questions in the Answer Sheets.

2.6a Indicate one reaction (write down the equation) for a process interfering in the sample analysis you have carried out in the presence of Ca2+.

2.6b A list of mistakes possible at different steps is given in the table in the answer 2– 2– sheet. Indicate which of the mistakes can lead to errors in CO3 and/or HPO4 content determination. Use the following symbols: “0” if no error is expected, “+”or “–“ if the result is higher (positive error) or lower (negative error) than the true one.

29 THEORETICAL EXAMINATION PROBLEMS

General Directions

- Write your name and code number on each page of the answer sheet. - You have 5 hours to fulfil the task. Failure to stop after the STOP command may result in zero points for the task. - Write answers and calculations within the designated box. - Use only the pen and the calculator provided. - There are 18 pages of Problems (incl. Cover Sheet and Periodic Table) and 22 pages of Answer Sheet. - An English-language version is available. - You may go to the restroom with permission. - After finishing the examination, place all sheets including Problems and An- swer Sheet in the envelope and seal. - Remain seated until instructed to leave the room.

30 Constants and useful formulas

Gas constant R = 8.314 J×K–1×mol–1 23 –1 Avogadro constant NA = 6.022·10 mol Planck constants h = 6.626·10–34 J×s h = 1.055·10–34 J×s Speed of light c = 3.00·108 m×s–1

h Uncertainty relation D×D³xp 2 Gibbs energy of a condensed phase at G = pV + const pressure p Excess pressure caused by surface DPin = 2σ / r tension Relation between equilibrium constant RTKGln = -D o and Gibbs energy r DG = DG° + RT·ln Q Isotherm of a chemical reaction product ofc (products) with Q = product ofc (reactants)

æöEA Arrhenius equation kA = expç÷- èøRT

V(cylinder) = pr2h S(sphere) = 4pr2 4 V(sphere) = pr3 3

31 Problem 1. PROTON TUNNELING 7 points Proton tunneling through energy barriers is an important effect, which can be observed in many complex species containing hydrogen bonds (DNA, proteins, etc.). Propanedial (malonaldehyde) is one of the simplest molecules for which intramolecular proton transfer can occur.

1.1.1 Draw the condensed formula of propanedial and the structures of two of its isomers, which can exist in equilibrium with propanedial.

1.1.2 In a water solution propanedial is a weak acid, its strength being comparable with that of acetic acid. Specify the acidic hydrogen atom. Explain its acidity (choose one version in the Answer Sheet).

On the plot below an energy profile of the intramolecular proton transfer is given (the dependence of energy on the distance of proton motion (in nm)). Energy curve has a symmetric double-well form. Energy, arb. units Energy,

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 L Distance, nm R

1.2.1 Draw the structures corresponding to two minima on this curve.

A proton is delocalized between two atoms and oscillates between two minima L and R with an angular frequency w = 6.48×1011 s–1. Probability density for a proton de- pends on time as follows:

32 1 2éù 22 22 Y(,)xtxxxxt= YLR() +Y () +( Y LR () -Y ()cos) () w , 2ëû

wavefunctions YL ()x and YR ()x describe a proton localized in the left and right wells, respectively:

Y2 2 2 Y Y L R

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 Distance, nm

1.3.1 Write down the expressions for the probability density at three moments: (a) t = 0, (b) t = p/(2w), (c) t = p/w. Sketch the graphs of these three functions.

1.3.2 Without calculations, determine the probability of finding the proton in the left well at t = p/(2w).

1.3.3 How much time is required for a proton to move from one well to another? What is the proton mean speed during the transfer?

1.3.4 From the energy curve, estimate the uncertainty of the position of proton form- ing hydrogen bonds. Estimate the minimal uncertainty of the proton speed. Compare this value with that obtained in 1.3.3 and draw a conclusion about the proton tunneling (choose one of the versions in the Answer Sheet).

33 Problem 2. NANOCHEMISTRY 8 points Metals of the iron subgroup are effective catalysts of hydrogenation of СО (Fischer- Тropsch reaction) Fe, Co CO + 3H2 CH4 + H2O

Catalyst (e.g. cobalt) is often used in the form of solid nanoparticles that have a spherical structure (fig.1). The reduction in size of the catalyst increases catalytic activity significantly. The unwanted side-reaction however involves the oxidation of the catalyst:

Co(s) + H2O (gas) CoO(s) + H2 (gas) (1)

Solid cobalt oxide (bulk) is formed in the reaction vessel. This causes an irreversible loss of the catalyst’s mass. Solid cobalt oxide can also be deposited on the surface of Co(s). In this case the new spherical layer is formed around the surface of the catalyst (see figure 2) and the catalytic activity drops.

Let us see how formation of nanoparticles affects the equilibrium of reaction (1). Useful equation: 2s GrGV00( ) = (bulk)+ r

0 2.1.1 Calculate the standard Gibbs energy DrG (1) and the equilibrium constant for the reaction (1) at T = 500 K.

34 2.1.2 Calculate the equilibrium constant for reaction (1) when the cobalt catalyst is dispersed in the form of spherical particles (Fig.1) of radius (a) 10–8 m,× (b) 10–9 m. The surface tension at the Co-gas interface is 0.16 J/m2. CoO forms a bulk phase.

The mixture of gases involved in the Fischer-Tropsch reaction (CO, CН4, Н2, Н2O) was put into a reaction vessel containing the cobalt catalyst. The total pressure is р = 1 bar, temperature is T = 500 K. The mole fraction of hydrogen (%) in the mixture is 0.15%.

2.2.1 At what minimum mole fraction of water (%) in the gas mixture the unwanted spontaneous oxidation of the catalyst becomes possible so that solid bulk CoO may appear in the system? Assume that cobalt catalyst is in the form of (a) a bulk phase

(b) spherical nanoparticles with ra = 1 nm (Fig. 1).

2.2.2 What would you suggest to protect Co nanoparticles from the spontaneous

oxidation with the formation of bulk CoO at a constant ratio pp(H22 O)/ (H ) and a constant temperature:

(a) to increase ra;

(b) to decrease ra;

Assume now that solid cobalt oxide forms a spherical layer around a nanoparticle of cobalt. In this case the nanoparticle contains both a reactant (Co) and a product

(CoO) (fig. 2). In the following problems denote surface tensions as sCoO-gas, sCoO-Co, radii as ra, rb, molar volumes as V(Co); V(CoO).

2.3.1 Write down the expression for the standard molar Gibbs function of CoO.

2.3.2 Write down the expression for the standard molar Gibbs function of Co.

35 Hint. If two spherical interfaces surround a nanoparticle, the excess pressure at its centre is given by the expression

ss12 PPinex-DD+D+ = P = PP 12 = 2 2 rr12

ri,si are radius and surface tension at the spherical interface i, respectively.

0 2.3.3 Express the standard Gibbs energy of the reaction (1) DrG(1,,) rr ab in terms of

0 sCoO-gas, sCoO-Co, ra, rb, V(Co); V(CoO) and DrG (1) .

2.3.4 When spontaneous oxidation of Co begins the radii of two layers in the

00 nanoparticle (Fig. 2) are almost equal, ra = rb = r0, and DDrabr0Grr(1,,)= Gr(1,). As-

sume that ssCoO-gas = 2 CoO-Co . Which plot in the Answer Sheet describes correctly the

0 dependence of Dr0Gr(1,) on r0 ?

2.3.5 What would you choose to protect Co nanoparticles from the spontaneous

formation of the external layer of CoO at a constant ratio pp(H22 O)/ (H ) and a constant temperature:

a) increase r0

b) decrease r0

c) change of r0 has no effect.

Reference data:

Substance ρ, g/cm3 o ΔfG 500 , kJ/mol Co (s) 8.90 CoO (s) 5.68 –198.4

H2O (gas) –219.1

36 Problem 3. UNSTABLE CHEMICAL REACTIONS 7 points Many chemical reactions display unstable kinetic behavior. At different conditions (concentrations and temperature) such reactions can proceed in various modes: stable, oscillatory or chaotic. Most of these reactions include autocatalytic elementary steps.

Consider a simple reaction mechanism involving autocatalytic step:

B+ 2X ¾k ¾®1 3X XDP+ ¾k ¾®2 (В and D are reagents, X is an intermediate and P is a product).

3.1.1 Write down the overall reaction equation for this two-step mechanism. Write the rate equation for X.

3.1.2 Deduce a rate equation using steady-state approximation. Find the orders: (i) a partial reaction order with respect to B; (ii) a partial reaction order with respect to D; (iii) the overall order of a reaction.

Let the reaction occur in an open system where reagents B and D are being continu- ously added to the mixture so that their concentrations are maintained constant and equal: [B] = [D] = const.

3.2.1 Without solving the kinetic equation draw the kinetic curve [X](t) for the cases:

1) [X]0 > k2/k1; 2) [X]0 < k2/k1.

3.2.2 Without solving the kinetic equation draw the kinetic curve [X](t) for the case when the reaction proceeds in a closed vessel with the initial concentrations: [B]0 =

[D]0, [X]0 > k2/k1.

Much more complex kinetic behavior is possible for the reactions with several intermediates. Consider a simplified reaction mechanism for cold burning of ethane in oxygen:

37 k1 C26 H+ X + ... ¾ ¾® 2X k X+¾¾®+ Y 2 2Y ...

k3 C26 H+ Y + ... ¾ ¾® 2P Under specific conditions this reaction displays oscillatory behavior.

Intermediates are peroxide C2H6O2 and aldehyde C2H4O, P is a stable product.

3.3.1 Identify X, Y, and P. Fill the blanks in the reaction mechanism.

Behavior of unstable reactions is often controlled by temperature which affects the rate constants. In the above oxidation mechanism oscillations of concentrations are possible only if k1 ³ k2. Parameters of the Arrhenius equations were determined experimentally:

3 –1 –1 –1 Step A, cm ×mol ×s EA, kJ×mol 1 1.0×1011 90 2 3.0×1012 100

3.4.1 What is the highest temperature at which oscillatory regime is possible? Show your calculations.

38 Problem 4. DETERMINATION OF WATER BY FISCHER TITRATION 8 points Determination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also

an excess of SO2 and pyridine (C5H5N, Py) – Fischer reagent. The following reactions occur during the titration:

SO2 + CH3OH + H2O + I2 = 2HI + CH3OSO3H Py + HI = PyH+I– + – Py + CH3OSO3H = PyH CH3OSO3

Iodine content is usually expressed in mg of water reacting with 1 mL of the titrant solution (hereunder T, mg/mL), which equals the mass of water (mg) reacting with 1.00 mL of the iodine solution. T is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water. In all calculations please use the atomic masses accurate to 2 decimal points.

4.1 Sometimes titration of water is performed in pyridine medium without methanol.

How would the reaction of I2 with SO2 and H2O occur in this case? Write down bal- anced reaction equation.

Calculate the T values of iodine solution in each of the following cases:

4.2.1 12.20 mL of Fischer reagent solution were used for titration of 1.352 g of so- . dium tartrate dihydrate Na2C4H4O6 2H2O.

4.2.2 A known amount of water (21.537 g) was placed into a 1.000 L volumetric flask which was filled by methanol up to the mark. For titration of 10.00 mL of the obtained solution, 22.70 mL of Fischer reagent solution were needed, whereas 2.20 mL of Fisher regent solution were used for titration of 25.00 mL of methanol.

39 4.2.3 5.624 g of water were diluted by methanol up to a total volume of 1.000 L (solution A); 22.45 mL of this solution were used for titration of 15.00 mL of a Fischer reagent (solution B).

Then 25.00 mL of methanol (of the same batch as used for the preparation of solution A) and 10.00 mL of solution B were mixed, and the mixture was titrated by the solution A. 10.79 mL of the latter solution were spent.

4.3 An inexperienced analyst tried to determine water content in a sample of CaO using Fischer reagent. Write down the equation(s) of reaction(s) demonstrating possible sources of errors.

For the titration of 0.6387 g of a hydrated compound Fe2(SO4)3·xH2O, 10.59 mL of iodine solution (T = 15.46 mg/mL) were consumed.

4.4.1 What other reaction(s), beside those given in the problem, can occur during the titration? Write down the equations of two such processes.

4.4.2 Write down an equation of the overall reaction of Fe2(SO4)3·xH2O with the Fischer reagent.

4.4.3 Calculate the composition of the hydrate Fe2(SO4)3·xH2O (x = integer).

40 Problem 5. A MYSTERIOUS MIXTURE (ORGANIC HIDE-AND-SEEK GAME) 7.5 points An equimolar mixture X of three colorless organic liquids A, B, C was treated by water with a drop of hydrochloric acid at heating to give, after separation from water, a 1:2 (molar ratio) mixture of acetic acid and ethanol without any other components. To the mixture after hydrolysis a catalytic amount (one-two drops) of concentrated sulfu- ric acid was added, and after long reflux (boiling with reflux condenser) a compound D, a volatile liquid with pleasant smell, was formed in 85% yield. Compound D is not identical to any of A, B, C.

5.1.1 Draw the structure of compound D.

5.1.2 Which class of organic compounds does D belong to? Choose the proper variant from those given in the Answer Sheet.

5.1.3 Even if the reflux is continued twice as long, the yield of D would not exceed 85%. Calculate the expected yield of D if 1:1 (molar ratio) mixture of ethanol and acetic acid is taken. Assume that: a) volumes do not change during the reactions; b) all concomitant factors, such as solvent effects, non-additivity of volumes, variation of temperature, etc. are negligible. If you cannot make a quantitative estimate, please indicate whether the yield will be: a) the same (85%); b) higher than 85%; c) lower than 85%.

1H NMR spectra of compounds A, B, C look very similar and each shows singlet, triplet and quartet with the ratio of integral intensities equal to 1:3:2.

The same mixture X was subjected to alkaline hydrolysis. A remained unchanged, and was separated. The remaining solution after acidification and short boiling gave 2:3 (molar ratio) mixture of acetic acid and ethanol with evolution of gas.

The mixture X (3.92 g) was dissolved in diethyl ether and underwent hydrogenation in the presence of Pd on charcoal catalyst. 0.448 L (standard conditions) of hydrogen were absorbed, but after the reaction A and C were isolated unchanged (3.22 g

41 of mixture were recovered) while neither B, nor any other organic compounds except diethyl ether could be identified after hydrogenation.

5.2.1 Determine and draw the structures of A, B, and C. 5.2.2 Which intermediate compounds are formed during the acidic hydrolysis of C, and basic hydrolysis of B.

The reaction of either B or C with acetone (in the presence of a base) with subsequent acidification by dilute HCl at gentle heating gives the same product, senecioic acid (SA), a compound widely occurring in Nature. Alternatively, senecioic acid can be obtained from acetone by treating it with concentrated HCl and subsequent oxidation of the intermediate product by iodine in alkaline solution. The latter reaction gives, besides sodium salt of senecioic acid, a heavy yellow precipitate E (see the scheme 2).

1. Me2CO/base B or C SA (1) 2. HCl, t C5H8O2 1. HCl cat. O SA (sodium salt) + E (2) 2. I2, NaOH

5.3.1 Determine the structure of senecioic acid and draw the reaction scheme leading to SA sodium salt from acetone.

5.3.2 Give the structure of E.

42 Problem 6. SILICATES AS THE BASE OF THE EARTH CRUST 7 points Silica and compounds derived from it, silicates, constitute ca. 90 % of the Earth crust substances. Silica gives rise to a beautiful material – glass. Nobody knows exactly how glass was discovered. There is a well-favored story related to Phoenician sailors who fused occasionally sea sand and soda ash. It is likely that they discovered the

secret of “liquid glass” (LGL) – sodium metasilicate (Na2SiO3) soluble in water.

6.1.1 The solution of LGL was used earlier as office glue. Write down the net ionic equation accounting for the ability of LGL to set in air.

Hydrolysis of LGL in water allows obtaining a colloidal solution of silicic acid.

6.1.2 Complete the Table in the Answer Sheet. Write down the net ionic equations matching the processes enumerated in the Table. For each process check the “Yes” box if it leads to changes of pH. Otherwise check the “No” box.

The structure of species occurring in aqueous solutions of silicates is rather complex. However, it is possible to distinguish the main building block of all species – ortho- 4– silicate tetrahedron (SiO4 , 1):

(1) n– For [Si3O9] ion found in aqueous solution of silicates:

6.2.1 Determine the charge (n).

6.2.2 Determine the number of oxygen atoms bridging adjacent tetrahedra.

6.2.3 Depict its structure joining together several tetrahedra (1). Take into account that any adjacent tetrahedron shares one vertex.

m– Charged monolayers with the composition [Si4O10] are found in kaolinite (clay).

43 6.2.4 Using the same strategy as in 6.2.1-6.2.3, depict a fragment of the layered structure joining 16 tetrahedra (1). Note that 10 tetrahedra have shared vertices with 2 neighbors each, and the rest 6 have shared vertices with 3 neighbors each.

Being placed into the LGL solution, salts of transition metals give rise to fancy “trees” tinted relevant to the color of the salt of the corresponding transition metal. Crystals of CuSO4·5H2O produce “trees” of blue color, whereas those of NiSO4·7H2O form green “trees”.

6.3.1 Determine the pH of 0.1 M aqueous solution of copper sulfate at 25°С, assuming that its hydrolysis occurs in small degree only. Use the value of the first acidity 2+ I –7 constant of [Cu(H2O)4] Ka =1·10 M.

6.3.2 Write down equation of a reaction between aqueous solutions of CuSO4 and sodium metasilicate (LGL). Take into account the pH values of aqueous solutions of the salts.

44 Problem 7. ATHEROSCLEROSIS AND INTERMEDIATES OF CHOLESTEROL BIOSYNTHESIS 7.5 points Cholesterol is a lipid wide-spread in living nature. Disruption of its metabolism leads to atherosclerosis and related potentially fatal diseases.

Substances Х and Y are two key intermediates of cholesterol biosynthesis in ani- mals. Х is an optically active monocarbonic acid composed of atoms of only three elements. It is formed in organisms from (S)-3-hydroxy-3-methylpentanedioyl- coenzyme A (HMG-CоА). This reaction is catalyzed by enzyme Е1 (which catalyses two types of reactions) and does not involve water as a substrate. Х is further me- tabolized into Х1 through a three-stage process requiring enzymes E2, E3, E4, which catalyze reactions of one and the same (and only one) type. Finally, Х1 spontaneously (non-enzymatically) decomposes to give isopentenyl pyrophosphate (3- methylbut-3-enyl diphosphate, IPP) and inorganic products:

- HO S E1 E2, E3, E4 O O O CoA P P X X1 * O O O OHO O- O- * HMG-CoA Scheme 1 IPP

7.1.1 In the Answer Sheet, choose the reaction type(s) for Е1 and Е3.

7.1.2 Draw the structure of X with stereochemical details and indicate absolute configuration (R or S) of the stereocenter.

Y is an unsaturated acyclic hydrocarbon. Its reductive ozonolysis leads to a mixture of only three organic substances Y1, Y2 and Y3 in a molar ratio of 2:4:1. Y is formed as a result of a number of successive coupling reactions of two isomeric substances: IPP and dimethyl allyl pyrophosphate (3-methylbut-2-enyl diphosphate, DAP) with subsequent reduction of a double bond in the final coupling product Y5. Carbon atoms IPP and DAP involved in the formation of C–C bonds during biosynthesis of Y are marked with asterisks.

45 O O O- P P * O O O- O- DAP

7.2.1 Write down the overall reaction equation for reductive ozonolysis of DAP, if dimethyl sulfide is used as the reducing agent.

The product of the final coupling reaction (hydrocarbon Y5) is formed when two hydrocarbon residues (R) of intermediate Y4 are combined:

O O +2H 2ROPOPO- Y5 RR O- O- Y4 Y 2PPi Scheme 2

At each coupling stage but that shown in Scheme 2, pyrophosphate is released in a molar ratio of 1:1 to the coupling product.

7.2.2 Determine molecular formula of Y, if it is known that Y2 and Y3 contain 5 and 4 carbon atoms, respectively.

7.2.3 Calculate the number of IPP and DAP molecules needed to give Y5, if it is known that all carbon atoms of isomeric pyrophosphates are incorporated into Y.

7.2.4 Draw the product of coupling reaction of one IPP molecule with one DAP molecule (C–C bond can be formed only by carbon atoms marked with asterisks), if it is known that subsequent reductive ozonolysis of the product of the coupling reaction gives Y1, Y2 and one more product, the latter containing phosphorus.

The only double bond reduced in Y5 during its metabolism into Y was formed in the reaction described in Scheme 2. All double bonds in Y and Y4 exist in trans configuration.

7.2.5 Draw the structures of Y and Y4 with stereochemical details.

46 Problem 8. ATRP ALLOWS NEW POLYMERS 8 points ATRP (Atom Transfer Radical Polymerization) is one of the most promising novel approaches towards polymer synthesis. This modification of radical polymerization is based on a redox reaction of organic halides with complexes of transition metals, Cu (I) in particular. The process can be described by the following scheme (M – monomer, Hal – halogen):

kact R-Hal+Cu(+)Hal(Ligand) . (2+) k R +Cu Hal2(Ligand)k

kdeact kP +M

kact (+) . (2+) R-M-Hal+Cu Hal(Ligand)k R-M +Cu Hal2(Ligand)k

kdeact

...

kp +(n-1)M

kact (+) . (2+) R-Mn-Hal+Cu Hal(Ligand)k R-Mn +Cu Hal2(Ligand)k kdeact

k . . t R-My +R-Mx R-M(y+x)R

The reaction rate constants are: kact – all activation reactions, kdeact – all reversible deactivation reactions, kp – chain propagation, and kt – irreversible termination.

8.1.1 Write down the expressions for the rates of ATRP elementary stages: activation (vact), deactivation (vdeact), propagation (vp) and termination (vt). Write down gen- eralized equation assuming just one reacting species R’X, where R’ means any of R- or R-Mn- and X means Hal.

47 Consider that the total number of polymeric chains is equal to that of initiator molecules. Assume that at each moment throughout polymerization all chains are of the same length.

8.1.2 Compare the rate of deactivation to the rates of ATRP elementary stages.

Dependence of monomer concentration ([M]) on reaction time (t) for ATRP is:

æö[]M × lnç÷ =-×× kp [] Rt, èø[]M0 · [M]0 – the initial monomer concentration, kp – the rate constant of propagation, [R ] – the concentration of active radicals.

To prepare a polymer sample by using ATRP, catalytic amounts of CuCl, organic ligand (L) and 31.0 mmol of monomer (methylmethacrylate, or MMA) were mixed. The reaction was initiated by adding 0.12 mmol of tosyl chloride (TsCl). Polymeriza- –1 –1 tion was conducted for 1400 s. kp is 1616 L·mol s , and the steady state concentration of radicals is 1.76·10–7 mol·L–1.

CH3 CH3 H2C H3C O O O H2C O CH3 OSi(CH3)3 SO2Cl

MMA TsCl HEMA-TMS

8.2.1 Calculate the mass (m) of the polymer obtained.

In another experiment the time of MMA polymerization was changed, all the rest reaction conditions being the same. The mass of the obtained polymer was 0.73 g. Then 2-(trimethylsilyloxy)ethyl methacrylate, HEMA-TMS (23.7 mmol) was added to the mixture and polymerization was continued for another 1295 s. MMA and HEMA- TMS reactivities are the same under reaction conditions.

8.2.2 Calculate the degree of polymerization (DP) of the obtained polymer.

48 8.2.3 Depict the structure of the obtained polymer (including end groups), showing MMA and HEMA-TMS units as A and B, respectively. If necessary, use the symbols in the copolymer structure representation: block (block), stat (statistical), alt (alternat-

ing), grad (gradient), graft (grafted). For example, (A65-graft-C100)-stat-B34 means that chains of polymer C are grafted on units A in the statistic copolymer of A and B.

ATRP was applied to synthesize two block copolymers, P1 and P2. One block in both block-copolymers was the same and was synthesized from mono-(2- chloropropionyl)-polyethylene oxide used as a macroinitiator: O

O CH3 H3C 58O

The other block in P1 consisted of styrene (C), and in P2 of p-chloromethylstyrene (D) units.

1H NMR spectra of the macroinitiator, P1 and P2 are given below. Integral intensities of characteristic signals can be found in the table.

8.3.1 Assign 1H NMR signals to the substructures given in the Answer Sheet.

8.3.2 Determine the molar fractions of units C and D and the molecular masses of P1 and P2.

8.3.3 Write down all possible reactions of activation occurring during the synthesis of P1 and P2. You may use R symbol to depict any unchanged part of the macromole- cule, but you should specify what substructure you use it for.

8.3.4 Draw the structure of P1 and one of possible structures of P2 representing poly(ethylene oxide) chain by a wavy line and showing units of co-monomers as C and D, respectively.

49 50 ANSWER SHEETS, SOLUTIONS, AND GRADING SCHEMES

PRACTICAL PROBLEMS

Problem 1. ION-EXCHANGE CHROMATOGRAPHY OF AMINO ACIDS

Question 1a 1b 1c 2-3 4 Total Points

Marks 9 0 3 72 2 86 20

Number of the given mixture of amino acids ______(A number between 301 and 600)

1.1a Draw the profile of coloration intensity on the plate sketch. 1.1b Indicate changes of eluents by drawing lines between the corresponding wells. 1 2 3 4 5 6 7 8 9 10 11 12

1.1c Labels of wells corresponding to the chosen fractions. Peak number Labels of wells 1 2 3

51 1.2-1.3 Content (in mg) of each amino acid in the amino acid mixture you were given. Your work

Complete the table. Peak Volume of Amino acid Wavelength Absorbance Amino acid mass number combined (3-letter l, nm Al in the given mix- fractions, mL code) ture, mg 1 2 3

ATTENTION. The print-offs with the spectra of your samples should be put into the envelope and delivered at the end of examination alongside with the An- swer Sheets.

1.4 Resonance structures of the substance responsible for the mixture coloration.

52 Solution and grading scheme

1.1a 3 marks for each peak if at least two blank wells between peaks 9 marks maximum.

1.1b Not graded.

1.1.c 1 point for each proper choice. Combined mixtures should include all fractions identified as “2” and “3” and should be free of fractions without amino acids. Frac- tions identified as “1” may be or may be not included. 3 marks maximum.

1.2-1.3 A × n ×V × M Content of an amino acid = l W , e ×l

Al is the absorbance of the sample calculated from the spectra, l is the optical length (1.0 cm), n is the dilution factor determined as a ratio of the aliquot of analyzing solution (0.1 mL) and the final volume of the sample in the cuvette, V is the volume of the combined fraction from the corresponding peak, and MW is the molar mass of the amino acid. 3 marks for correct formula, 3 marks for correct determination of optical densities (1 mark for each assay) 1 mark penalty for miscalculation.

Content of each amino acid is determined by using the following plot (values recalcu- lated from volumes reported by students and absorbance values recorded by spectrophotometer)

53 Marks

P max

min min max max m m m m % mass determined acc exp exp acc

Scoring:

m < min macc OR m > max macc 0 marks

(m - min macc ) min macc < m < min mexp Pmax marks (min mexp - min macc )

min mexp < m < max macc Pmax marks

Min Min Max Max

Amino acid macc/mexact mexp/mexact mexp/mexact macc/mexact Pmax ·100% ·100% ·100% ·100% Cys 70% 85% 110% 115% 22 His 60% 85% 110% 115% 22 Arg 25% 45% 65% 85% 22

72 marks maximum

1.4 As it is given in the task text, mixed disulfide and 2-thio-5-nitrobenzoic acid are formed in the reaction.

54 NO2

- O O O - - O S O - + O NH3 + H N S + - OH- 3 - O S S + O S O SH - -H2O O O NO2 O NO2 (I) (II)

NO2

Under slightly alkaline conditions, thiol group of (II) dissociates, and thiophenolate- anion is formed. Resonance structures can be realized for this compound:

- O - - - COO COO O - - O O O + - + + N S N S N S - - O O O ,

The electronic structure of asymmetrical disulfide (I) does not differ considerably from that of the original Ellmann reagent. Therefore, it can be concluded that the compound responsible for coloration is thiophenolate-anion (a form containing C=S bond).

2 marks for three correct ionized structures, one of which contains C=S bond 1 mark for three other structures, all without C=S bond 1 mark for less than three structures, one of which contains C=S bond. 2 marks maximum.

55 Problem 2. DETERMINATION OF CARBONATE AND HYDROGEN PHOSPHATE IN AN ABRASIVE SAMPLE

Question 1a 1b 2 3 4 5 6a 6b Total Points

Marks 25 5 25 25 5 5 1 9 100 20

2.1a Standardization of NaOH solution

Titration No Initial burette reading, Final burette reading, Volume of NaOH solu- mL mL tion consumed (V1), mL 1 2 3

Final volume of NaOH solution (V1,f), mL

2.1b Calculation of NaOH concentration Your work

с(NaOH) = ______mol/L

2.2 The first titration of the sample (BCG)

Titration No Initial burette reading, Final burette reading, Volume of NaOH solution mL mL consumed (V2), mL 1 2 3

Final volume of NaOH solution (V2,f), mL

56 2.3 The second titration of the sample (TP)

Initial burette reading, Final burette reading, Volume of NaOH solution Titration No mL mL consumed (V3), mL 1 2 3

Final volume of NaOH solution (V3,f), mL

2– 2.4 Calculation of the mass of CO3 Your work

2– m(CO3 ) = ______g

2– 2.5 Calculation of the mass of HPO4 Your work

2– m(HPO4 ) = ______g

Additional questions 2.6a Indicate one reaction (write down the equation) for a process interfering in the sample analysis you have carried out in the presence of Ca2+.

57 2.6b A list of mistakes possible at different steps is given in the table. Indicate which 2– 2– of the mistakes can lead to errors in CO3 and/or HPO4 content determination. Use the following symbols: “0” if no error is expected, “+”or “–“ if the result is higher (positive error) or lower (negative error) than the true one. Error Mistake Step HPO 2– CO 2– content 4 3 content

Incomplete removal of CO2 1 Too large excess of K C O on calcium precipi- 2 2 4 2 tation Too late indication of the endpoint (overtitration) 3 on NaOH standardization Insufficient washing of the filter at CaC O filtra- 2 4 4 tion Overtitration of the sample against BCG 5 Overtitration of the sample against TP 6

H2CO3: pKa1 = 6.35; pKa2 = 10.32

H2C2O4: pKa1 = 1.25; pKa2 = 4.27

Extra samples given or column broken Column broken Student’s Lab assistant’s Problem No Sample No and replaced signature signature

Solution and grading scheme

2.1a, 2.2, 2.3 The values of the final volumes V1,f, V2,f, and V3,f, (as reported in the Answer Sheet) are graded according to the following scheme:

DV = |Vreported – Vtrue|,

Vreported is either V1,f, V2,f, and V3,f, Vtrue is the corresponding master value (will be given with the copies of students’ works).

58 Value of DV Marks

DV £ DVexpected 25

DV ³ DVacceptable 0

æöDVVacceptable -D DVexpected £ DV £ DVacceptable 25´ ç÷ èøDVVacceptable -D expected

The values of DVexpected and DVacceptable (in mL) are listed in the table below. DV, mL expected acceptable

V1,f 0.10 0.25

V2,f 0.15 0.40

V3,f 0.15 0.40

25 marks maximum for each titration

2.1b Calculation of NaOH concentration c(HCl) ´V (HCl) ´V (aliquot) 1.214 (mol / L)´10.00(mL) ´10.00(mL) c(NaOH) = = = V ( flask) ´V (NaOH ) 100.0(mL)´V1, f (mL)

2– 2.4 Calculation of the mass of CO3

2– 2– c(NaOH ) ´ (V1, f -V3, f )´V ( flask) m(CO3 ) (g) = M(CO3 )´1/2´ = V (aliquot) c(NaOH ) (mol / L) ´ (V -V ) (mL)´100.0 (mL) = 60.01 (g/mol)´1/2´ 1, f 3, f ´ 0.001 (L / mL) = 10.00 (mL)

2– 2.5 Calculation of the mass of HPO4

2– 2– c(NaOH ) ´ (V3, f -V2, f ) ´V ( flask) m(HPO4 ) (g) = M(HPO4 )´ = V (aliquot) c(NaOH ) (mol / L) ´ (V -V ) (mL) ´100.0 (mL) = 95.98 (g/mol)´ 3, f 2, f ´ 0.001 (L / mL) = 10.00 (mL)

2.1b, 2.4, 2.5 Correctness of calculations is graded by:

59 1) comparing the numerical result (including the number of significant figures) reported by a student with that obtained from the student's data using correct method of calculation; 2) checking out the a student's way of calculation.

5 marks maximum for completely correct calculation of each value

Penalty for errors in calculations or data presentation. Penalty marks (for cal- Error type culation of each value)

1 Incorrect method of calculation –5 2 Mistakes in dilution factor –1 3 Confusion between units of measurement –1 4 Mistake in stoichiometric (equivalency) relationships –1 5 More or less than 4 significant figures in c(NaOH) –0.5* More than 4 or less than 3 significant figures in 6 * 2– 2– –0.5 m(CO3 ) and m(HPO4 ) Rounding errors affecting the 1st or 2nd significant fig- 7 –0.5** ures in the final result

8 Miscalculations and slips –0.2 per error

*Only the number of significant figures in the final answer is taken into account. **Not applied if the error originates from insufficient number of significant figures in previously calculated c(NaOH). No double penalty!

2.6a 2+ – + Ca + H2PO4 ® CaHPO4 + H or 2+ 2– + 3Ca + 2HPO4 ® Ca3(PO4)2 + 2H In course of these processes free protons evolve influencing the results of titration. 1 mark if the Answer Sheet contains at least one correct reaction equation. Incorrect equations are not penalized.

60 2.6b Error 2– 2– Mistake CO3 HPO4 content content

Below pH 4.5 (first endpoint, V2) H2CO3 is not titrated

Between pH 4.5 and 10 (second endpoint, V3) H2CO3 is – + titrated

Thus, increase of V3 ; V1 and V2 unchanged 2– No influence, C2O4 is too weak base 0 0

Increase of V1 and proportional decrease of c(NaOH)

[as c(NaOH)V1 is equivalent to the initial amount of stan- – – dard HCl and so remains constant]; V2 and V3 unchanged

Losses of sample leads to proportional decrease of V2 + – and V3 and therefore V3-V2; V1 unchanged

Increase of V2, V1 and V3 unchanged 0 –

Increase of V3, V1 and V2 unchanged – +

61 THEORETICAL PROBLEMS

Problem 1. PROTON TUNNELING

Question 1.1 1.2 2.1 3.1 3.2 3.3 3.4 Total Points

Marks 3 3 2 4.5 2 4 6 24.5 7

1.1.1 Structures:

Propanedial

1st isomer

2nd isomer

1.1.2 Circle the acidic hydrogen atom

O O

C C H C H H H

The acidity of propanedial is caused by a) the stability of a carbanion due to conjugation with two carbonyl groups b) weakness of C–H bond in a carbonyl group c) hydrogen bonds between two propanedial molecules

The correct answer ______

62 1.2.1 The structures corresponding to minima on energy curve:

1.3.1 The probability density (a) t = 0 Y2 (x ,0) =

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 Distance, nm

(b) t = p/(2w)

2 æöp Y ç÷x, = èø2w

63 Y2

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 Distance, nm

2 æöp Y ç÷x, = èøw

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 Distance, nm

1.3.2

The probability of finding the proton in the left well = ______

1.3.3 The time of proton transfer Your work: t =

64 The proton mean speed Your work:

v =

1.3.4 The uncertainty of proton position Dx =

The minimal uncertainty of proton velocity Your work:

Dv =

a) Proton is a rather heavy particle, and its tunneling in malonaldehyde can be described in classical terms of position and velocity b) Proton tunneling is a purely quantum effect; it cannot be described in classical terms c) Uncertainty of proton velocity is so large that tunneling cannot be observed experimentally d) Uncertainty of proton velocity is so small that tunneling cannot be observed experimentally

The correct answer is ______

Solution and grading scheme

1.1.1 The structures of propanedial and two of its isomers

O=CH-CH2-CH=O 1 mark

65 H O O C C H C H H 1 mark OH C H H C C H O 1 mark 3 marks maximum

1.1.2

Acidic hydrogen atom is in CH2 (in enol forms acidic hydrogen is in OH). 1 mark

Acidity of СН2 group is caused by the stability of carbanion due to conjugation with two carbonyl groups. The first answer is correct. 2 marks 3 marks maximum

1.2.1 The distance between two minima on the energy curve is 0.06 nm. In a purely aldehyde form H O

H O such distance between two possible positions of proton is impossible. Tunneling takes place only in enol Z-form:

H H O O O O C C C C H C H H C H H H 1 mark for each structure 2 marks maximum

1.3.1 Expressions and plots of probability density 1 (а) Y2(,0)x= éù Y 22222() xxxxx +Y () +Y () -Y () = Y () 1 mark 2ëûLRLRL The probability density is concentrated in the left well:

66 2 Y L

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 L Distance, nm R 0.5 marks (b) In the middle of the time interval

2æöp1 22 Yç÷x, = éù YLR () xx +Y () 1 mark èø22w ëû The probability density has a symmetric form, a proton is delocalized between two wells:

2 2 Y Y ( L + R )/2

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 L Distance, nm R 0.5 marks

2æöp1 22222 (c) Yç÷x, = éù Y+Y-Y+YYLRLRR () xxxx () () () = () x 1 mark èøw2ëû The probability density is concentrated in the right well:

2 Y R

-0,06 -0,04 -0,02 0,00 0,02 0,04 0,06 L Distance, nm R 0.5 marks 4.5 marks maximum

67 1.3.2 The probability of finding the proton in the left well is 1/2, because probability function is symmetric, and both wells are identical. 2 marks

1.3.3 The time of transfer from one well to another is t = p / w. 3.14 t = = 4.8510× -12 s. 2 marks 6.48× 1011 The proton velocity: 0.06× 10-9 V = = 12 m/s. 2 marks 4.85× 10-12 4 marks maximum

1.3.4 The uncertainty of proton position is approximately equal to half of the distance between minima, that is 0.03 nm (0.06 nm was also accepted). 1 mark The minimal uncertainty of velocity can be obtained from the uncertainty relation: h 1.055× 10-34 D»V = = 1000 m/s. 3 marks 0.001 2mxD2× ×× 0.03 10-9 6.02× 1023 Comparing this uncertainty with the velocity 12 m/s we see that the notion of proton velocity during transfer from one well to another is senseless. Therefore, proton tunneling is a purely quantum phenomenon and cannot be described in classical terms. The second conclusion is correct. 2 marks 6 marks maximum

68 Problem 2. NANOCHEMISTRY

Question 1.1 1.2 2.1 2.2 3.1 3.2 3.3 3.4 3.5 Total Points

Marks 1 2 4 2 1 5 2 3 2 22 8

2.1.1 Thermodynamic data for the reaction (1): Your work:

0 ΔrG (1) =

K =

2.1.2 Equilibrium constant for the reaction (1) with cobalt nanoparticles: Your work:

(a) K (r = 10–8 m) = (b) K (r = 10–9 m) =

69 2.2.1 Minimum water content in the mixture: Your work:

(a) H2O%(bulk Co) =

–9 (b) H2O%(nanoparticles with r = 1×10 m) =

2.2.2 The correct answer is (mark the proper box):

(a) (b) (c)

2.3.1 Standard molar Gibbs function of CoO (external layer)

0 Gr(CoO,)b =

2.3.2 Standard molar Gibbs function of Co (internal layer):

0 G(Co,,) rrab=

2.3.3 Standard Gibbs energy for the reaction (1) with the double-layered nanoparticles

0 DrG(1,,) rr ab =

70 0 2.3.4. Plot ΔrG (1,r0) vs. r0

(b) ) 0 r )

(a) 0 (1, 0 (1,r G 0 r D G r D

r 0 r0

(d)

) (c) 0 ) 0 G(1,r r D G(1,r r D

r r 0 0

The correct plot is (mark the proper box):

(a) (b) (c) (d)

2.3.5 The correct answer is (mark the proper box):

(a) (b) (c)

Solution and grading scheme

2.1.1 The Gibbs energy and the equilibrium constant of reaction (1)

71 000 DrffGGG500(1) = D ,500 (CoO,s) -D ,500 (H 2 O,g) = -+198.4 219.1 = 20.7 kJ/mol 0.5 marks

DG0 (1) 20700 -r 500 - Kee = RT = 8.314 × 500 = 6.8810×-3 0.5 marks 1 mark maximum

2.1.2 The standard Gibbs energy of the reaction (1) with the spherical cobalt

nanoparticles of radius ra is

o 0 000 DrGrGGGG500(1, a ) = bulk ,500 (CoO,s) +-- 500 (H 2 ,g) 500 (H 2 O,g) sph (Co) =

0000æö2sCo-gasV (Co) = GGGG500 (CoO,s)+- 500 (H 2 ,g) 500 (H 2 O,gas) -+ç÷ 500 (Co,s) = èøra

o2sCo-gasV (Co) =D-rG500 (1); ra M 10-63× 59.0 m V (Co) = Co = = 6.610× -6; r(Co) 8.90 mol

–8 –9 for spherical particles with ra = 10 , 10 m one gets, respectively 2s V (Co) Co-gas = 210 and 2100 J/mol. ra o DraGr500 (1,) is equal to 20.5 (a), and 18.6 (b) kJ/mol, respectively. The equilibrium constant is calculated from the equation

æöD Gro (1,) Kr(1, ) = expç÷- ra500 ; a ç÷RT èø

--38 --39 Krr(1,aa ) = 7.22´ 10 ; = 10 m Krr(1,aa ) = 11.4´ 10 ; = 10 m 2 marks maximum

2.2.1 The standard Gibbs energy for reaction (1) involving nanoparticles of cobalt is

oo2sCo-gas DrGrGV500(1, ar ) = D- 500 (1) (Co) ra o o DrG500 (1) is 20.7 kJ/mol. For spherical cobalt particles with ra = 1 nm DraGr500 (1,) is 18.6 kJ/mol. Solid cobalt oxide can be formed spontaneously when Gibbs energy of reaction (1) is negative. The inequality for bulk cobalt is:

72 ooæöpp(H22 ) æö (H O) DD+D-£rrrG(1) = G500 (1) RT ln ç÷ = G 500 (1) RT ln ç÷ 0 èøpp(H22 O) èø (H )

and for spherical cobalt nanoparticles with ra = 1 nm: 2s ooæöpp(H22 )Co-gas æö (H O) DD+D--£rararG(1, r ) = G500 (1, r ) RT ln ç÷ = G 500 (1) V (Co) RT ln ç÷ 0 èøp(H22 O) rpa èø (H )

p(H O) The minimum ratios 2 are 145.6 (a) and 87.7 (b), respectively. p(H)2 The hydrogen pressure is 1 bar×0.0015 = 1.5×10–3 bar The minimum pressures of water are 1.5×10–3×145.6 = 0.218 bar (a) and 1.5×10–3×87.7 = 0.132 bar (b), for the bulk cobalt and for nanoparticles, respectively.

–9 H2O%(bulk Co) = 21.8% H2O%(nanoparticles with ra = 1*10 m) = 13.2%. We assume that bulk cobalt oxide is formed. 4 marks maximum

2.2.2 For the spontaneous oxidation 2s oCo-gas æöp(H2 O) DD--£rarG(1,) r= G500 (1)(Co)ln0 V RT ç÷ rpa èø(H)2 and 2s oCo-gas æöp(H2 O) D-£rG500 (1) V (Co) RT ln ç÷ rpa èø(H)2 The left hand side of the last inequality becomes more positive with the increase of

ra. At certain point the inequality will be disturbed and the spontaneous oxidation will not take place. So, to protect cobalt nanoparticles from the spontaneous oxidation in

this case one has to lengthen the radius ra. The answer (a) is correct. 2 marks maximum

2.3.1 The equation for the standard molar Gibbs function of CoO (external layer) reads:

0 22ssCoO-gaso CoO-gas GrGVGVsph(CoO, b ) = bulk (CoO)++ (CoO) = (CoO,s) (CoO) rrbb 73 1 mark maximum

2.3.2 The equation for the standard molar Gibbs function of Co (internal layer) reads:

0 æö2sCoO-gas 2sCoO-Co GrrGVsph(Co,ab , ) = bulk (Co)++ (Co) ç÷ = èørrba 2s oæöCoO-gas 2sCoO-Co = GV(Co,s)++() Co ç÷ èørrba The expression in brackets gives the additional pressure in the internal layer (see the Hint). 5 marks maximum

2.3.3 The standard Gibbs energy for reaction (1) with the double-layered nanoparticles is

000oo DrG(1, r a , r b ) = G sph (CoO, r b ) +-- G (H22 , gas ) G (H O,gas) G sph ( Co , r a , r b ) = oooo = GGgasGG(CoO,s)+ (H22 , ) - (H O,gas) -+ (Co,s)

2ssCoO-gasæö CoO-gas s +VV(CoO) -+ 2 (Co) ç÷CoO-Co = rbèø rr ba

o2sCoO-gas 2sCoO-Co = D+-rG(1)( VV (CoO) (Co)) - V (Co) rb ra 2 marks maximum

2.3.4 Under the assumptions made 2s oo0 CoO-gas 2sCoO-Co DrabrrGrrGrG(1, , ) = D (1,0 ) = D+ (1)( V (CoO) -- V (Co)) V (Co) = rrba

o2sCoO-gas æö3 = D+-rG(1)ç÷ VV (CoO) (Co) r0 èø2 The term in brackets in the right-hand side is positive

æö3 –6 3 ç÷VV(CoO)- (Co) = 6.56×10 m èø2

0 æö1 DrGr(1,)0 is directly proportional to ç÷. The plot (a) is correct. èør0 3 marks maximum

2.3.5. The spontaneous forward reaction (1) is possible, when DrG(1,r0) £ 0, and

74 2sCoO-gas æö3pHO D+G0 (1) V (CoO) -£ V (Co) RT ln 2 r rpç÷2 0Hèø2 The term in brackets in the left-hand side is positive. The left hand side of the ine-

quality becomes more positive with the decrease of r0. At certain point the inequality will be violated and the spontaneous oxidation will not take place.

In order to protect nanoparticles from oxidation in this case one has to shorten the radius r0. The answer (b) is correct. 2 marks maximum

75 Problem 3. UNSTABLE CHEMICAL REACTIONS

Question 1.1 1.2 2.1 2.2 3.1 4.1 Total Points

Marks 2 4.5 4 3 3 3 19.5 7

3.1.1 The overall reaction equation

The kinetic equation for X d[X] = dt

3.1.2 The rate equation Your work:

d[P] = dt

Reaction orders:

with respect to B (i): ______with respect to D (ii): ______overall (iii): ______

76 3.2.1 1) An open system, [X]0 > k2/k1

[X]

[X] 0

2) An open system, [X]0 < k2/k1

[X]

[X] 0

77 3.2.2 A closed system, [B]0 = [D]0, [X]0 > k2/k1

[X]

[X] 0

3.3.1 X –

Y –

P –

C26 H++® X ... 2X

X+®+ Y 2Y ...

C26 H++® Y ... 2P

3.4.1 The highest possible temperature:

Your work:

T =

78 Solution and grading scheme

3.1.1 The overall reaction equation B + D ® P 1 mark The kinetic equation for X d[X] = kk[B][X]2 - [D][X] 1 mark dt 12 2 marks maximum

3.1.2 Under the steady-state conditions d[P] = kk[D][X] = [B][X]2 , dt 21 whence k [D] [X] = 2 k1[B]

d[P] k 22[D] = 2 3 marks dtk1[B] The reaction order is 2 with respect to D, –1 with respect to B; the overall order is 1 0.5 marks for each correct order 4.5 marks maximum

3.2.1 In an open system the initial reaction rate is: d[X] = [B][X](kk [X]- ) dt 12

1) If [X]0 > k2/k1, then d[X]/dt > 0 at any time, and the concentration of X mono- tonically increases:

[X]

t 2 marks

79 2) If [X]0 < k2/k1, then d[X]/dt < 0 at any time, and the concentration of X mono- tonically decreases. Two types of kinetic curves are possible. If k2/2k1 < [X]0 < k2/k1, the inflection point will be observed on the kinetic curve:

[X]

But if the initial concentration is too low, [X]0 < k2/2k1, the concentration will mono- tonically decrease together with its derivative.

[X]

(any of these curves was accepted as a right answer) 2 marks 4 marks maximum

3.2.2 In a closed system the initial reaction rate is:

d[X] 2 = k100200 [B] [X]- k [D] [X] = [B] 00102 [X]( kk [X] ->) 0 dt t=0 Hence, at the beginning of the reaction [X] increases but it cannot increase infinitely and finally goes to its initial value, because the second reaction is irreversible:

80 [X]

2 marks for maximum 1 mark for the asymptotic value 3 marks maximum

3.3.1 X – C2H6O2, Y – C2H4O, P – C2H6O. Dots denote O2 and H2O.

CH26+ CHO 262 +®O 2 2CHO 262

CHO262+®+ CHO 24 2CHO 24 HO2 CH++® CHO 2CHO 26 24HO2 26

0.5 marks for each unknown substance (X, Y, P, three blanks) 3 marks maximum

3.4.1 At the highest possible temperature the rate constants are equal:

æöEEAA,1 æö ,2 AA12exp ç÷-- = exp ç÷ èøRT èø RT EE- T = AA, 2 ,1 = 354 K A Rln 2 A1 1 mark for the calculation 2 marks for the correct answer 3 marks maximum

81 Problem 4. DETERMINATION OF WATER BY FISCHER TITRATION

Question 1 2.1 2.2 2.3 3 4.1 4.2 4.3 Total Points

Marks 1 1.25 1.75 2.25 1 2 1 2.25 12.5 8

4.1. Equation:

4.2.1. Calculation of the T value: Your work:

T = ______mg/mL

4.2.2. Calculation of the T value: Your work:

T = ______mg/mL

4.2.3. Calculation of the T value: Your work:

T = ______mg/mL

4.3. Equation(s):

82 4.4.1 Equation(s):

4.4.2. Equation:

4.4.3. The composition of the crystallohydrate is: Your work:

Formula of the salt Fe2(SO4)3·xH2O: x = ______

Solution and grading scheme

4.1 Equation:

I2 + SO2 + 2 H2O + 4 Py = 2 PyHI + (PyH)2SO4 1 mark (0.75 marks for not accounting for the formation of Py salts)

4.2.1 T is equal to: . M(Na2C4H4O6 2H2O) = 230.05 2M(H2O) = 36.04

m(H2O) = 1.3520 · 36.04 / 230.05 = 0.2118 g = 211.8 mg 1 mark for the formula T = 211.8 / 12.20 = 17.36 mg/mL 0.25 marks for the correct result (within 2 digits) 1.25 marks maximum

83 4.2.2 T is equal to:

Volume of iodine spent for 10 mL of pure CH3OH = 2.20·10.00 / 25.00 = 0.88 mL (0.5 marks for the correct formula of pure methanol titration) T = 21.537×0.01×103 / (22.70 – 0.88) = 9.87 mg/mL

More accurately 10.00 mL of the solution contains (1000-21.5)×10.00 / 1000 = 9.785 mL of methanol

Volume of iodine spent for 9.785 mL of pure CH3OH = 2.20·9.785 / 25.00 = 0.86 mL T = 21.537×0.01×103 / (22.70 – 0.86) = 9.86 mg/mL (1 mark for the formula of water titration, only 0.5 marks without subtracting 0.88)

T = 9.87 mg/mL 0.25 marks for the correct result 1.75 marks maximum

4.2.3 T is equal to: Approach 1.

Let 1 mL of CH3OH contain x mg H2O, then 1 mL of A contains ((1.000 – 0.006)·x +

5.624) mg H2O. 15.00·T = 22.45·(0.994·x + 5.624) – 1st titration, 10.00·T = 25.00·x + 10.79·(0.994·x + 5.624) – 2nd titration. Hence, x = 1.13 mg/mL, T = 10.09 mg/mL (10.10 without taking into account 0.994 factor)

Approach 2.

Let y mL of B be spent for the titration of water, contained in 1 mL of CH3OH. Then 22.45·5.624 10.79·5.624 T = (1st titration) = (2nd titration). 15.00 - 22.45·0.994·y 10.00 - 25.00y -10.79y Hence, y = 0.1116 and T = 10.10 mg/mL

T = 10.09 mg/mL (10.10 without taking into account 0.994 factor) (2 marks for the correct formulas (with or without taking into account 0.994 factor) and 0.25 marks for the correct result (10.10 or 10.09))

84 2.25 marks maximum

4.3 Equation(s):

CaO + SO2 = CaSO3

2CaO + 2I2 = CaI2 + Ca(OI)2

6CaO + 6I2 = 5CaI2 + Ca(IO3)2

(Instead of CaO, Ca(OH)2 may be written.) 1 mark for ANY correct equation 1 mark maximum

4.4.1 Equation(s):

Fe2(SO4)3 + 2HI = 2FeSO4 + I2 + H2SO4 1 mark

Fe2(SO4)3 + H2O + SO2 + CH3OH = 2FeSO4 + CH3OHSO3 + H2SO4 1 mark (or in ionic form) 2 marks maximum

4.4.2 Equation:

Fe2(SO4)3·xH2O + (x–1)I2 + xSO2 + xCH3OH =

= 2FeSO4 + xCH3OHSO3 + H2SO4 + 2(x–1)HI 1 mark maximum

4.4.3. The composition of the crystallohydrate is:

M(Fe2(SO4)3·xH2O) = 399.9 + 18.02x 0.6387·18.02x mH2O (g) = ; 1 mark (399.9+18.02x) x mH2O (g) = 10.59(mL)´15.46(mg/mL)´0.001(g/mg)´ 1 mark x -1

0.1637·(399.9 + 18.02x) = 11.51x – 11.51; x = 8.994 . Formula: Fe2(SO4)3 9 H2O x = 9 0.25 marks (for correct answer) 2.25 marks maximum

85 Problem 5. A MYSTERIOUS MIXTURE (ORGANIC HIDE-AND-SEEK GAME)

Question 1.1 1.2 1.3 2.1 2.2 3.1 3.2 Total Points

Marks 5 5 10 30 10 10 5 75 7.5

5.1.1 Structure of product D

5.1.2 Which class of organic compounds does D belong to? Check the appropriate box. Note! Only one checkmark is allowed. Several checkmarks will lead to 0 marks for this question.

ketones ethers acetals esters alcohols aldehydes glycols o o o o o o o

5.1.3 The expected yield of D

The yield is equal to 85% o; lower than 85% o; greater than 85% o

Your work:

yield = %

5.2.1 The structures of A, B, and C.

A B C

86 5.2.2 Draw in the boxes intermediate compounds formed during the acidic hydrolysis of C, and basic hydrolysis of B.

+ H /H2O t C CH3COOH + C2H5OH

– – OH /H2O OH /H2O – B CH3COO + C2H5OH

5.3.1 The structure of senecioic acid and the reaction scheme leading to SA sodium salt from acetone.

5.3.2 The structure of E.

Solution and grading scheme

5.1.1 Structure of product D O

H3C O CH3 Ethyl acetate, ethyl ethanoate 87 Any structural formula or any shorter versions (CH3COOC2H5) including the adopted shortcuts for organic radicals (Me, Et, Ac), or systematic IUPAC name – 5 marks

5.1.2 Which class of organic compounds does D belong to? Check the appropriate box. ketones ethers acetals esters alcohols aldehydes glycols o o o þ o o o The only correct answer is ester – 5 marks

5.1.3 The expected yield of D Statement that the reaction is an equilibrium without any further actions – 1 mark Answer to quantitative question: lower than 85% – 2 marks

Qualitative estimation of yield can be done assuming that the reaction is at equilibrium, and that the equilibrium constant is supposed to not vary with temperature and composition of the reaction mixtures. [AcOEt][H O] (0.85)2 K = 2 = = 4.2 [AcOH][EtOH] 0.15× 1.15 Calculation of yield using this constant in 1:1 mixture gives 67% yield = 67% maximum – 10 marks, if the yield is computed within 67±1% error limits

5.2.1 The structures of A, B, and C. OEt COOEt OEt OEt COOEt OEt HC≡COEt CH2(COOEt)2 CH3C(OEt)3 ethoxyacetylene, ethynylethyl ether diethyl malonate triethyl orthoacetate, 1,1,1- triethoxyethane A B C Each structure represented by structural formula or unambiguous linear notation – 10 marks. Systematic name given in place of structure – 5 marks 30 marks maximum

88 5.2.2 Draw in the boxes intermediate compounds formed during the acidic hydrolysis of C, and basic hydrolysis of B. a) Malonic acid is formed as intermediate in the hydrolysis of diethyl malonate – 5 marks

COOEt + COOH H /H2O t CH3COOH + C2H5OH COOEt COOH – CO2

C Also accepted will be monoethyl malonate – 2 marks. Maximum 5 marks b) Hydrolysis of ethoxyacetylene starts from the addition of hydroxide to the triple bond to give unstable enolic form of ethylacetate, into which it immediately is transformed

OEt – OH O OH–/H O OH /H2O 2 – CH3COO + C2H5OH OEt OEt B

Indication of any of keto- or enol forms of ethylacetate – 5 marks Hydrolysis of strong ether bond to give hydroxyacetylene, or any forms coming along this path (ketene, diketene) is impossible and is not allowed – 0 marks Maximum for a) and b) together – 10 marks

5.3.1 The structure of senecioic acid From acetone alone the synthesis includes aldol condensation, dehydration, with subsequent iodoform reaction 3 marks

O O O H+ H+ I /OH– 2 O 2 HO -H O 2 OH

Senecioic acid structure alone – 4 marks, with scheme – maximum 10 marks

5.3.2 The structure of E.

Iodoform, triiodomethane, CHI3 – 5 marks

89 Problem 6. SILICATES AS THE BASE OF THE EARTH CRUST

Question 1.1 1.2 2.1 2.2 2.3 2.4 3.1 3.2 Total Points

Marks 3 9 2 2 3 10 5 3 37 7

6.1.1 The net ionic equation accounting for the ability of LGL to set in air

6.1.2 Write down the net ionic equations matching the processes enumerated in the Table. For each process check the “Yes” box if it leads to changes of pH. Otherwise check the “No” box. а) protonation of ortho-silicate ions leading to the formation of Si-OH groups Reaction equation:

Yes No 4– b) formation of hydrated [SiO4(H2O)2] anions Reaction equation:

Yes No c) polycondensation of ortho-silicate ions leading to the formation of Si-O-Si bonds Reaction equation:

Yes No

n– 6.2 For [Si3O9] ion found in aqueous solution of silicates: 6.2.1 Determine the charge (n). Your justification

n = ______

90 6.2.2 Determine the number of oxygen atoms bridging adjacent tetrahedra. Your justification

Number of oxygen atoms = ______

6.2.3 Depict the ion structure joining together several tetrahedra (1).

6.2.4 The fragment of the layered structure joining 16 tetrahedra (1) Your justification

Structure

6.3.1 pH of 0.1 M aqueous solution of copper sulfate Your justification

pH = ______

91 6.3.2 Equation of a reaction between aqueous solutions of CuSO4 and sodium metasilicate (LGL)

Solution and grading scheme

6.1.1 2– – SiO3 + 2CO2 + 2H2O = “H2SiO3”¯ (Silica acid gel) +2HCO3 or 2– – SiO2(OH)2 + 2CO2 + H2O = “H2SiO3”¯ + 2HCO3 or 2– 2– SiO3 + CO2 + H2O = “H2SiO3”¯ + CO3 3 marks maximum 2 marks if silicic acid is written in any form 1 mark if the reaction contains carbonate (or bicarbonate) ions with silicic acid missing

6.1.2 а) protonation of ortho-silicate ions leading to the formation of Si-OH groups 4– 3– – SiO4 + H2O = [SiO3(OH)] + OH or 4– + 3– SiO4 + H = [SiO3(OH)] or 2– + – [SiO2(OH)2] + H = [SiO(OH)3] Yes No 4– b) formation of hydrated [SiO4(H2O)2] anions 4– 4– SiO4 + 2H2O = [SiO4(H2O)2] Yes No c) polycondensation of ortho-silicate ions leading to the formation of Si-O-Si bonds 4– 6– – 2 SiO4 + H2O = [O3Si-O-SiO3] + 2 OH or 4– + 6– 2 SiO4 + 2H = [O3Si-O-SiO3] + H2O or 2– 2– – 2SiO2(OH)2 + Н2O = [O-Si(OH)2-O-Si(OH)2-O] + 2 OH Yes No 9 marks maximum 2 marks for each correct reaction 1 mark for each correct choice of check-boxes

92 6.2.1 n = 6 (assuming oxidation numbers of silicon (+4) and oxygen (–2), or taking into account its structure and the charge of orthosilicate ion (–4)) 2 marks maximum 1 mark penalty for miscalculation

6.2.2

Si3O9 ≡ 3 [SiO4] – 3 O, i.е. there are 3 oxygen atoms bridging adjacent tetrahedra 2 points maximum 1 point penalty for miscalculation

6.2.3

3 marks maximum

6.2.4 Calculations: m = 4 (assuming oxidation numbers of silicon (+4) and oxygen (–2), or taking into account its structure and the charge of orthosilicate ion (–4))

Si4O10≡ 4[SiO4] – 6O, i.е. the formula of the tetrahedron is now SiO2.5, which is possible if 1 О atom belongs to this tetrahedron and the other three are shared between 2 tetrahedra (their contribution = 3/2). This is possible if the tetrahedra are set on a plane and joined together through all apexes of their bases.

10 marks maximum 2 marks for charge determination 3 marks for determination of the number of oxygen bridges

93 5 marks for the correct structure 1 mark penalty if 6 to 15 tetrahedra shown, the connection being correct 3 marks penalty if less than 6 tetrahedra shown (i.е. it is not clear that polyhedra form layer) 4 marks penalty for connection via apexes, but in 3D network 4 marks penalty for connection via apexes, but in 1D chain 0 mark of 5 for any other structure

6.3.1 рН = 4 2+ + + Cu(H2O)4 + H2O = Cu(OH)(H2O)3 + H3O , + I 1/2 –4 + [H ]≈ (c Ka ) =1·10 M, pH = –log[H ] = 4 5 marks maximum 1 mark penalty for miscalculation + I 2 marks penalty for wrong expression of [H ] via Ka 2 marks penalty if there are some errors in definition of рН (e.g. using ln instead of log) 3 marks penalty if the hydrolysis reaction is wrong

6.3.2

СuSO4 + Na2SiO3 + 2H2O = Cu(OH)2¯ + “H2SiO3”¯ + Na2SO4 or

2СuSO4 + Na2SiO3 + 2H2O = (CuOH)2SO4¯ + “H2SiO3”¯ + Na2SO4 This (or those) reaction(s) (apart from formation of copper silicate) can be deduced from the fact that the reaction describes mutual (self-amplifying) hydrolysis. It comes from the previous parts of the task: рН of LGL is greater than 7 (see questions 6.2), and рН of copper sulfate solution is less than 7 (see 6.3.1). 3 marks maximum 2 marks if reaction coefficients are wrong

1 mark if only one of two principal precipitates shown (Cu(OH)2¯ or “H2SiO3”¯)

94 Problem 7. ATHEROSCLEROSIS AND INTERMEDIATES OF CHOLESTEROL BIOSYNTHESIS

Question 1.1 1.2 2.1 2.2 2.3 2.4 2.5 Total Points

Marks 12 12 5 12 7 8.5 16 72.5 7.5

7.1.1 A number of reaction types is listed in the table below. All reactions involved in metabolism of HMG-CoA to IPP are in the list. Choose those types of reactions which are catalyzed by Е1 and Е3 (put numbers in appropriate places). No Reaction type 1. Dehydration 2. Decarboxylation 3. Dephosphorylation 4. 4 electron reduction 5. Release of the reduced form of coenzyme A (CoA-SH) 6. Monophosphorylation Oxidation of hydroxyl group as the third stage of HMG-CoA β-oxidation 7. cycle

Е1 ______

Е3 ______

7.1.2 Draw the structure of X with stereochemical details and indicate absolute configuration (R or S) of the stereocenter.

7.2.1 Write down the overall reaction equation for reductive ozonolysis of DAP with dimethyl sulfide used as the reducing agent.

95 7.2.2 Determine molecular formula of Y.

Your justification

Number of carbon atoms______

Number of hydrogen atoms_____ Molecular formula:______

7.2.3 Calculate the number of IPP and DAP molecules needed to give Y5. Your justification:

Number of IPP molecules ____ Number of DAP molecules ____

7.2.4 Draw the product of coupling reaction between one IPP molecule and one DAP molecule, subsequent reductive ozonolysis of which gives Y1, Y2 and one more product, the latter containing phosphorus.

7.2.5 Draw the structures of Y and Y4 with stereochemical details.

96 Y

Solution and grading scheme

7.1.1 Е2-Е4 catalyze one and the same (and only one) reaction type. The only reaction which can be carried out three times in a row is monophosphorylation (all the rest reaction types are not consistent with either initial or final products). This is also supported by presence of pyrophosphate residue in IPP and liberation of inorganic products (including inorganic phosphate) upon spontaneous decomposition of Х1.

Х is a monocarboxylic acid composed of atoms of three elements: carbon, hydrogen and oxygen. It can contain neither sulfur which is found in CoA nor phosphorus which is introduced into intermediates on the pathway from HMG-CoA towards IPP or present in CoA. Thus, Е1 catalyzes non-hydrolytic removal of CoA from HMG- CoA and is not involved in phosphorylation. Since water is not a substrate in this reaction, liberation of CoA must be conjugated with another reaction which affects the carboxylic group esterified in HMG-CoA. The only possible variant is its 4 electron reduction towards hydroxyl group. Е1 can not catalyze dehydration because of optical activity of Х (removal of water leads to the loss of sole chiral center). Decarboxy- lation is excluded, since Х, being an acid, must contain a carboxylic group. Oxidation of tertiary hydroxyl group in HMG-CoA according to β-oxidation mechanism is impossible. Further evidence comes from the fact that the carboxylic group initially involved in thioester bond formation is present as the residue of hydroxyl group in IPP. So:

Е1 4, 5

Е3 6

12 marks maximum E1: 9 marks if 2 variants are given and both are correct. 4 marks if only one variant is given and it is correct

97 4 marks if two correct and one incorrect variants are given 0 mark if one correct and any number of incorrect variants are given 0 mark if more than three variants are given. Е3: 3 marks if only correct variant is given. Otherwise 0 mark

7.1.2 Based on the reaction types catalyzed by Е1 and configuration of HMG-CoA stereocenter, the structure of Х is: HO HOOC (R) OH Х, mevalonic acid

Note the absolute configuration of the chiral center is changed as a result of HMG- CoA metabolism into mevalonic acid due to alteration of substituents priority.

CoA HO S E1 HO HOOC HOOC (R) (S) O OH 12 marks maximum 8 marks for correct structural formula 4 marks for correct stereochemistry (only in case both the structure is correct and R- isomer is indicated; otherwise 0 mark). No penalty for incorrect stereochemistry or absence of stereochemical information

7.2.1 Reaction equation for reductive ozonolysis

O- - H O O O H3C CH3 P P + O + H3C CH3 + O P O + S O O 3 S O - - O O P O- O O - O O O

5 marks maximum 4.5 marks for correct products (1.5 marks each), incorrect structures not penalized 0.5 mark for correct equation coefficients

7.2.2 DAP molecule contains only one carbon atom which can be involved in the formation of С–С bond during Y biosynthesis. Irrespective of the way this molecule is incorporated in Y, ozonolysis of this fragment will lead to dimethyl ketone (acetone). (See DAP ozonolysis reaction in 7.2.1). Thus, acetone can be unambiguously attrib- 98 uted to Y1, since it contains 3 carbon atoms (Y2 and Y3 contain 5 and 4 carbon atoms, respectively). Taking into account the ratio between ozonolysis products, we have:

nY(C)= 2×nY1(C) + 4×nY2(C) + nY3(C) = 2×3 + 4×5 + 4 = 30 Y is an acyclic molecule, thus DAP residues can be found only at its ends. Y has only two ends, since IPP contains only two elongation sites (at least three such sites are needed to get a branched molecule). Since reductive ozonolysis of one Y molecule produces two acetone molecules, Y contains 30 carbon atoms.

To determine the number of hydrogen atoms double bonds in Y should be counted. Formation of each double bond reduces by 2 the number of hydrogen atoms in the coupling product as compared to the sum of atoms of starting substances. The ratio of Y to the sum of its ozonolysis products is 1:7 (2+4+1), which corresponds to 6 double bonds in Y. Then, by using the general formula for alkanes we have:

n(H) = 2×nY(C) + 2 – 2×nc=c = 30×2 + 2 – 6×2 = 50

Y (squalene) formula – С30Н50. Number of carbon atoms 30 Number of hydrogen atoms 50

Gross formula of Y – С30Н50 12 marks maximum 8 marks for correct justification of molecular formula 4 marks for correct molecular formula

7.2.3 IPP and DAP are structural isomers containing 5 carbon atoms each. Since all carbon atoms of these substances are found in Y, one can calculate the total quantity of IPP and DAP molecules needed to synthesize Y:

n(IPP&DAP) = nY(C) / 5 = 30 / 5 = 6 The number of DAP molecules was determined earlier and is equal to 2. Then, 4 molecules of IPP are needed. Number of DAP molecules 2 Number of IPP molecules 4 7 marks maximum 3.5 marks for calculation of the total number of DAP and IPP molecules 3.5 marks for correct individual numbers of DAP and IPP molecules

99 7.2.4 All possible combinations that do not alter hydrocarbon skeleton are given below (pyrophosphate fragments not shown). Two groups of products differing in carbon atoms involved in coupling reaction are separated by the dashed line. IPP fragments should be attached to DAP so that ozonolysis of the product leads to Y2 containing 5 carbon atoms. Only one variant is possible if stereochemistry is not taken into consideration and two variants with stereochemical details

No * * * + + No No

Yes

(E) O- O P O- O O P O- O or

-O O- O P P O O- (Z) O O

The upper isomer is geranyl pyrophosphate 8.5 marks maximum 8.5 marks for correct structure No penalty for stereochemistry, any correct structure is accepted 2.5 marks if ozonolysis of the product leads to acetone, but does not lead to the compound with 5 carbon atoms 2.5 marks if ozonolysis of the product leads to the compound with 5 carbon atoms, but does not lead to acetone 0 mark for any other variant

100 7.2.5 It is seen from the coupling reaction (Scheme 2) that Y4 contains 15 carbon atoms or 1 DAP and 2 IPP fragments, the latter being attached to the former con- secutively. It is important to note that Y3 can not be found in two hydrocarbon residues originating for Y4, since Y3 is formed as a result of ozonolysis in a molar ratio of 1:1 to Y. Thus, geranyl phosphate is the intermediate on the way to Y (all double bonds in trans configuration). Attachment of the next IPP fragment to geranyl phosphate leads to the product giving 1 molecule of Y1 and 2 molecules of Y2 upon its ozonolysis. Thus, Y4 structure with stereochemical details:

(E) (E) O- O P O- O O P O- O Y4, farnesyl pyrophosphate

Combining two hydrocarbon fragments of Y4 and taking into account that the double bond between them is being reduced we get the following structure of Y:

(E) (E) (E) (E)

Y, squalene 16 marks maximum 9 marks for farnesyl pyrophosphate (6.5 marks for correct structural formula and 2.5 marks for correct stereochemistry) 7 marks for squalene (5 marks for correct structural formula and 2 marks for correct stereochemistry) 2.5 marks penalty for unreduced double bond in squalene

101 Problem 8. ATRP ALLOWS NEW POLYMERS

Question 1.1 1.2 2.1 2.2 2.3 3.1 3.2 3.3 3.4 Total Points

Marks 8 9 5 11 14 16.5 12 10 13.5 99 8

8.1.1 Expressions for the rates:

vact = vp =

vdeact = vt =

8.1.2 Compare rates using operators <<, ≤, », ≥, >>

vdeact vact vdeact vt

vdeact vp

8.2.1 Mass of the obtained polymer. Your justification:

m =

102 8.2.2 Degree of polymerization of the obtained polymer. Your justification:

DP =

8.2.3 Structure of the obtained polymer.

8.3.1 Fill in the right column with symbols (a-g) of 1H NMR signals corresponding to substructures in the left column.

* CH2 * O CH2

* H H

H H H

* H H

H H * * * H Cl * Cl HH

103 8.3.2 Composition and molecular weights of copolymers P1 and P2. Your justification: Your justification:

n(C) = n(D) = M(P1) = M(P2) =

8.3.3. All possible reactions of activation P1:

P2:

104 8.3.4 Structure of P1 and one of possible structures of P2 P1: P2:

Solution and grading scheme

8.1.1 Expressions for the rates of ATRP elementary stages: activation (vact), deacti-

vation (vdeact), propagation (vp) and termination (vt)

vact = kact·[R-Hal]·[CuHal(Ligand)k] 2 marks · vdeact = kdeact·[R ]·[CuHal2(Ligand)k] 2 marks · vp = kp·[R ]·[M] 2 marks · 2 vt = 2kt·[R ] 2 marks (no penalty for missing 2) 8 marks maximum

8.1.2 Comparison of rates of ATRP elementary stages Since all the chains grow with equal rate, the process proceeds as living polymerization. Living radical polymerization is possible only if concentration of active radicals is low to prevent chain transfer and termination. So:

vdeact >> vact 3 marks The portion of active radicals must be small, so the equilibrium is shifted towards dormant species.

vdeact >> vp 3 marks Propagation rate should be much slower than that of deactivation to make chains propagate with equal rate.

105 vdeact >> vt 3 marks Termination does not occur since the total number of polymer chains is equal to a certain constant number – number of initiator molecules. 9 marks maximum

8.2.1 Calculation of mass (m) of the obtained polymer. 1st variant

[M ] = [ M ]0 exp(-× kP [ Rt ] ) or n(MMA) = n0 (MMA)exp(-× kP [ Rt ] ) 1 mark Quantity of MMA monomer remaining after polymerization during 1400 s is 31.0 ×exp(-1616 ×1.76 ×10-7 ×1400) = 20.8 mmol. 2 marks Quantity of monomer consumed during polymerization: 31 – 20.8=10.2 mmol 1 mark Mass of the obtained polymer is mnM = D×× (MMA) (MMA) = (10.2/1000) 100.1 = 1.03g 1 mark

2nd variant

[M ] = [ M ]0 exp(-× kP [ Rt ] ) or n(MMA) = n0 (MMA)exp(-× kP [ Rt ] ) 1 mark Quantity of MMA monomer consumed during 1400 seconds of polymerization is

·-7 Dn(MMA) = n0 (MMA)(1 --×× exp( kp [ Rt ] )) = 31.0 ×-××× (1 1616 1.76 10 1400) = 10.2 mmol 3 marks Mass of the obtained polymer is mnM = D×× (MMA) (MMA) = (10.2/1000) 100.1 = 1.03g 1 mark

3rd variant

æö[]M -7 ln ç÷ = -kP [ Rt ×- ] = 1616 ××× 1.76 10 1400 = -0.398 1 mark èø[]M0 []M = e- 0.398 = 0.672 1 mark []M0 [Mn ] (MMA) = [Mn ]00 (MMA)

nn(MMA) = 0.672×0 (MMA) = 20.8 mmol 1 mark Quantity of monomer consumed during polymerization is 31 – 20.8 = 10.2 mmol 1 mark 106 Mass of the obtained polymer is mnM = D×× (MMA) (MMA) = (10.2/1000) 100.1 = 1.03g 1 mark m = 1.03 g 5 marks maximum

8.2.2 Calculation of degree of polymerization (DP) of the obtained polymer. The number of growing chains is equal to the number of TsCl molecules (0.12 mmol) 2 marks At the first stage, 7.3 mmol of MMA was consumed (0.73/100.1). The total quantity of monomers at the beginning of the 2nd stage is 23.7 + 23.7 = 47.4 mmol. 2 marks Since the monomers have the same reactivity, they will be involved in polymerization with the same rate. Quantity of monomers consumed during the second stage is

-7 Dn = n0 (1- exp(-k P [R×]t)) = 47.4(1- exp(-1616 ×1.76 ×10 ×1295)) = 14.6 mmol. 4 marks Totally 7.3+14.6 = 21.9 mmol of monomers was polymerized during two stages. 2 marks DP = 21.9/0.12 = 182.5 1 mark DP = 182-183 (all answers within this range are credited) 11 marks maximum

8.2.3 Structure of the obtained polymer. The product of polymerization is a block copolymer because it was obtained by se- quential polymerization on living chains. The first block is built of MMA units solely. The DP is 7.3/0.12 = 60.8 ≈ 61 monomer units. The second block is obtained by copolymerization of two competing monomers with the same reactivity. So, it is a statistical copolymer. Fractions of A and B in the 2nd block are equal because their concentrations in the reaction mixture at the beginning of the 2nd stage were equal. The DP of the 2nd block is 183-61 = 122 monomer units (121 is also correct if the total DP in 8.2.2 is 182).

Ts-A61-block-(A-stat-B)61-Cl or Ts-A61-block-(A61-stat-B61)-Cl 14 marks maximum 4 marks for block copolymer with blocks A and co-AB 4 marks for an indication of the statistic character of the 2nd block

107 1 mark for equal fractions of A and B in the 2nd block 2 marks for correct DPs of each block 1 mark for indication of terminal groups

8.3.1 Assignment of NMR signals to substructures given in the Answer Sheet.

* CH2 * a, b, g 3x1.5 marks O CH2

* H H c 2 marks H H H

* H H d 2 marks H H * * * e 4 marks H Cl * Cl f 4 marks HH 16.5 marks maximum

8.3.2 Determination of molar fractions of units C and D and molecular weights of P1 and P2. Intensity of multiplets b and g is 40.2, so intensity per 1 proton is 40.2 / 4 / 58 = 0.173 for both copolymer spectra 2 marks Intensity of multiplet с is 13.0, which is equivalent to 13.0 / 0.173 = 75 protons. Tak- ing into account that each styrene ring has 5 aromatic protons, DP of styrene block is 75/5 = 15. 2 marks Molar fraction of styrene units in P1 is 15 / (15+58) = 20.5% 1 mark Intensity of multiplet d is 10.4, which is equivalent to 10.4/0.173=60 protons. Since each monomer unit of p-chloromethylstyrene has 4 protons, DP of PCS is 60/4=15. 2 marks Molar fraction of D is 15/(15+58) = 20.5% 1 mark M(P1) = 15.03 + 58×44.05 + 72.06 + 15×104.15 + 35.45 = 4240 2 marks

108 M(P2) = 15.03 + 58×44.05 + 72.06 + 15×152.62 + 35.45 = 4967 2 marks 12 marks maximum

8.3.3 All possible reactions of activation occurring during the synthesis of P1 and P2. 10 marks maximum P1: (1.5+2) marks

O O

O CH3 O H C O Cu(+)Cl(Ligand) O Cu(+2)Cl (Ligand) 3 + k H3C CH 2 k 58 58 + Cl CH3

R R Cl CH Cu(+)Cl(Ligand) Cu(+2)Cl (Ligand) + k + 2 k

Here R is used for the macroinitiator fragment with one or several styrene units attached.

P2: (1.5+2+3) marks

109 O O

O CH3 O H C O Cu(+)Cl(Ligand) O Cu(+2)Cl (Ligand) 3 + k H3C CH 2 k 58 58 + Cl CH3

R R Cl CH

Cu(+)Cl(Ligand) Cu(+2)Cl (Ligand) + k + 2 k

Cl Cl

R R Cl Cl

Cu(+)Cl(Ligand) Cu(+2)Cl (Ligand) + k + 2 k

CH Cl 2

Here R is used for the macroinitiator fragment with one or several p- chloromethylstyrene units attached.

8.3.4 The structure of P1 and one of possible structures of P2 P1 is a block copolymer of PEO and PS. The PS block contains 15 units. P2 is a block copolymer composed of PEO block and branched styrene block. The integral intensity of multiplet f is 2.75, so 2.75 / 0.173 = 15.9, that is about 16 protons or 8 chloromethyl groups. d) If there is no branching in molecule P2, it would contain 15 chloromethyl groups. Each branching reduces the number of such groups by 1. Thus P2 has 15 – 8 = 7 branchings. Every structure with 7 branchings is correct if each monomer unit is linked with not more than 3 other monomer units

C C C C C C C C Cl Р1 R C C C C C C C

110 Р2

Cl Cl D Cl D Cl D Cl Cl D

D D Cl D

Cl R D Cl

D Cl D D Cl D Cl D D Cl

Cl D Cl

Cl 13.5 marks maximum 2 marks for P1 7.5 marks for completely correct structure of P2 4 marks for structure of P2 with nonzero but incorrect number of branchings 4 marks penalty if there is a unit linked with more than 3 other monomer units

111 STATISTICAL ANALYSIS OF THE PROBLEMS

Experimental problem 1

30 Works . 20

0 0-2 2-4 4-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20 Points

Experimental problem 2

30 Works . Works 20

0 0-2 2-4 4-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20 Points

Experiment total

80 70 60 50 40

Works . 30 20 10 0 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 Points

112 Theoretical problem 1

50 45 40 35 30 25 20 Works . Works 15 10 5 0 0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 6 6,5 7 Points

Theoretical problem 2

Works . Works 10

0 00,511,522,533,544,555,566,577,58 Points

Theoretical problem 3

70 60 50 40 30 Works . Works 20 10 0 00,511,522,533,544,555,566,57 Points

113 Theoretical problem 4

45 40 35 30 25 20

Works . Works 15 10 5 0 0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 6 6,5 7 7,5 8 Points

Theoretical problem 5

50 45 40 35 30 25 20 Works . Works 15 10 5 0 0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 6 6,5 7 7,5 Points

Theoretical problem 6

40 35 30 25 20

Works . Works 15 10 5 0 0 0,5 1 1,5 2 2,5 3 3,544,5 5 5,5 6 6,5 7 Points

114 Theoretical problem 7

Works . Works 20

0 0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 6 6,5 7 7,5 Points

Theoretical problem 8

35 30 25 20 15 Works . 10 5 0 00,511,522,533,544,555,566,577,58 Points

Theory total

45 40 35 30 25 20

Works . Works 15 10 5 0 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60 Points

115 Total results

40 35 30 25 20

Works . Works 15 10 5 0 0-5 5-10 10- 15- 20- 25- 30- 35- 40- 45- 50- 55- 60- 65- 70- 75- 15 20 25 30 35 40 45 50 55 60 65 70 75 80 Points

116 RESULTS PER STUDENT

Practical Theoretical Total Rank Name Country exam exam Medal (max 100) (max 40) (max 60)

1 Lei XU China 19.771 56.300 76.071 Gold 2 Yuan FANG China 25.768 47.541 73.309 Gold 3 Leonid ROMASHOV Russian Federation 32.556 39.712 72.267 Gold 4 Vasiliy VOROBYEV Russian Federation 28.200 43.408 71.609 Gold 5 Zi-yang ZHANG China 20.397 50.192 70.588 Gold 6 Ying Yu HO Chinese Taipei 23.385 47.016 70.401 Gold 7 Dimitry Ur'evic LOUTCHKO Germany 26.113 44.115 70.228 Gold 8 Jae Soo KIM Korea Republic 35.000 34.721 69.721 Gold 9 Simon GOURDIN-BERTIN France 27.457 40.529 67.986 Gold 10 Tae Gon OH Korea Republic 22.718 45.027 67.745 Gold 11 Wei-Lun HUANG Chinese Taipei 23.136 44.214 67.349 Gold 12 Dawid Grzegorz LICHOSYT Poland 37.806 27.458 65.264 Gold 13 Krzysztof Cezary KOSINSKI Poland 28.209 35.500 63.710 Gold 14 Le YANG China 13.729 49.136 62.865 Gold 15 Stanislav TEREHOV Russian Federation 25.214 37.564 62.778 Gold 16 Philipp Albert STEININGER Germany 31.457 30.621 62.078 Gold 17 Bavorn HONGSRICHINDA Thailand 32.136 29.501 61.638 Gold 18 Eugeny NEKHOROSHEV Russian Federation 12.991 48.481 61.472 Gold 19 Wojciech Dominik MAGON Poland 23.189 37.745 60.935 Gold 20 Chang Ho LEE Korea Republic 22.308 37.602 59.911 Gold 21 Hande BOYACI Turkey 26.934 32.959 59.893 Gold 22 Sumit SOMANI India 16.191 43.569 59.759 Gold 23 Soham MEHTA India 13.961 45.797 59.757 Gold 24 Martin LUKACISIN Slovakia 26.865 32.135 59.000 Gold 25 Gyula PALFY Hungary 19.153 39.712 58.866 Gold 26 Karolis LEONAVICIUS Lithuania 25.819 32.580 58.399 Gold 27 Minh Nguyen Thi NGOC Vietnam 11.198 47.043 58.241 Gold 28 Aurimas VYSNIAUSKAS Lithuania 23.883 34.212 58.095 Gold 29 Hyeonjin BAE Canada 26.640 31.362 58.003 Gold 30 Przemyslaw Krzysztof TREDAK Poland 19.577 38.006 57.582 Gold 31 Ehsan SHABANI Iran 28.640 28.783 57.423 Gold 32 Kai-Jui CHANG Chinese Taipei 17.192 40.002 57.194 Silver 33 Tanatorn KHOTAVIVATTANA Thailand 16.211 40.858 57.069 Silver 34 Samvel BARDAKHCHYAN Armenia 15.005 41.874 56.878 Silver 35 Erik ANDRIS Slovakia 28.207 28.495 56.701 Silver 36 Andres LAAN Estonia 18.755 37.569 56.324 Silver 37 Dzianis HRAMAZDOU Belarus 22.505 33.778 56.283 Silver 38 Frank Meng LIN Chinese Taipei 18.174 37.600 55.774 Silver 39 Ostap CHERVAK Ukraine 16.959 38.429 55.388 Silver 40 Brian Kihoon LEE United States 15.090 40.156 55.246 Silver 41 Ctirad CERVINKA Czech Republic 27.686 27.448 55.134 Silver 42 Andrew TULLOCH Australia 18.304 36.723 55.027 Silver 43 Justin KOH United States 25.441 29.480 54.921 Silver 44 Boris FACKOVEC Slovakia 22.485 32.361 54.846 Silver 45 Ha Phan Tran HONG Vietnam 12.005 42.435 54.439 Silver 46 Vikas PRAJAPATI India 13.814 40.031 53.845 Silver 47 Attila LOVAS Hungary 26.400 26.948 53.348 Silver

117 48 Narek DSHKHUNYN Armenia 4.345 48.776 53.121 Silver 49 Alexander PUCHHAMMER Austria 20.593 32.482 53.075 Silver 50 Matthew James CLIFFE United Kingdom 14.930 38.144 53.074 Silver 51 Zeqi YANG Singapore 16.398 36.193 52.591 Silver 52 Yerdos ORDABAYEV Kazakhstan 20.921 31.572 52.493 Silver Frederick Robert William Meath 53 United Kingdom 17.898 34.542 52.44 Silver MANNERS 54 Jonathan LIN New Zealand 26.423 25.257 51.679 Silver 55 Petr HOSEK Czech Republic 21.159 30.501 51.661 Silver 56 Pornchai KAEWSAPSAK Thailand 14.586 35.859 50.445 Silver 57 Assaf Avraham SHAPIRA Israel 19.610 30.628 50.238 Silver 58 Zhivko Atanasov GEORGIEV Bulgaria 26.508 23.691 50.199 Silver 59 Lorinc SARKANY Hungary 20.169 29.837 50.005 Silver 60 Stefan Michael PUSCH Germany 14.908 35.053 49.961 Silver 61 Teuku Mahfuzh Aufar KARI Indonesia 19.428 30.374 49.802 Silver 62 Vasile GRAUR Moldova 26.640 23.153 49.793 Silver 63 Bernardas MORKUNAS Lithuania 12.547 37.061 49.608 Silver 64 Jong Soo YOON Korea Republic 10.165 39.100 49.265 Silver 65 Kenneth BREWER United States 14.651 34.210 48.861 Silver 66 Muhamad FAIZ Indonesia 12.228 36.543 48.770 Silver 67 Ioan Teodor TROTUS Romania 12.031 36.713 48.744 Silver 68 Alena VASKOVA Belarus 17.046 31.173 48.219 Silver 69 Andi KIPPER Estonia 22.559 25.525 48.084 Silver 70 Stephan PRIBITZER Austria 25.128 22.485 47.613 Silver 71 Kartik RAMESH Australia 12.765 34.793 47.558 Silver 72 Vlad Alexandru PUSCASU Romania 16.199 31.328 47.527 Silver 73 Kaveh MATINKHOO Iran 15.477 31.912 47.388 Silver 74 Chuan Zheng LEE New Zealand 21.221 25.880 47.101 Silver Thais Macedo Bezerra Terceiro 75 Brazil 6.805 40.260 47.064 Silver JORGE 76 Balint BALAZS Hungary 21.647 25.160 46.807 Silver 77 Jan KOGOJ Slovenia 11.955 34.755 46.71 Silver 78 Vincenzo GRANDE Italy 15.437 31.064 46.501 Silver 79 Ionut Gabriel DUMITRU Romania 11.135 35.354 46.489 Silver 80 Linh Bui LE Vietnam 12.571 33.874 46.445 Silver 81 Christian OBERENDER Germany 14.951 31.469 46.420 Silver 82 Kainar KAMALOV Kyrgyzstan 13.044 33.033 46.077 Silver 83 Jan BITENC Slovenia 17.450 28.546 45.996 Silver 84 Benjamin CHEN Singapore 16.318 29.592 45.909 Silver 85 Matias Daniel Gomez ELIAS Argentina 15.446 30.324 45.771 Silver Oscar Carl Gunnar 86 Sweden 23.490 22.188 45.677 Silver GRANBERG 87 Techin CHULADESA Thailand 12.103 33.544 45.647 Silver 88 Gabriel Eduardo Sanoja LOPEZ Venezuela 26.834 18.351 45.185 Bronze 89 Mihails ARHANGELSKIS Latvia 7.063 37.904 44.967 Bronze 90 Christian MARBOE Denmark 21.828 23.029 44.857 Bronze 91 Derek Steven Hung-Che CHAN United Kingdom 10.785 34.012 44.797 Bronze 92 Takashi HIROI Japan 7.035 37.491 44.526 Bronze 93 Ayana BADRAKOVA Kazakhstan 16.875 27.297 44.172 Bronze 94 Volodymyr TKACHENKO Ukraine 13.031 30.852 43.883 Bronze Rafael de Cesaris Araujo 95 Brazil 11.694 32.007 43.701 Bronze TAVARES 96 Shabnam SHARIFZADEH Iran 15.455 28.165 43.620 Bronze 97 Florian LANGMANN Austria 11.869 31.661 43.529 Bronze 98 Raoul ROSENTHAL Netherlands 24.573 18.952 43.525 Bronze 99 Ahmet Selim HAN Turkey 14.401 28.972 43.373 Bronze

118 100 Naru TANAKA Japan 6.453 36.910 43.363 Bronze 101 Vincentius Jeremy SUHARDI Indonesia 17.791 25.311 43.102 Bronze 102 Jun Yan GOH Malaysia 25.190 17.864 43.054 Bronze 103 Shotaro TSUNODA Japan 7.017 35.679 42.696 Bronze 104 Taavi PUNGAS Estonia 14.557 28.097 42.654 Bronze 105 Arshavir GHAHRAMANYAN Armenia 0.900 41.616 42.516 Bronze 106 Kirill POLISCHUK Ukraine 8.118 34.367 42.485 Bronze 107 Gonzalo Jose MUNAR Argentina 20.343 22.106 42.449 Bronze 108 Guang Jun Joseph LIM Singapore 11.426 30.918 42.344 Bronze 109 Andrew Brian CAIRNS Ireland 18.006 24.284 42.290 Bronze 110 Lucia Domenica MEIER Switzerland 22.642 19.641 42.283 Bronze 111 Thorbjørn Juul MORSING Denmark 24.159 17.954 42.112 Bronze 112 Dauren KALIYEV Kazakhstan 18.433 23.598 42.031 Bronze 113 Tsimafei BALOTNIK Belarus 15.347 26.017 41.364 Bronze 114 Zijun (Jim) GE Australia 7.476 33.874 41.350 Bronze 115 Zhanbolat ZHOLGELDIYEV Kazakhstan 14.752 26.473 41.225 Bronze 116 Serdar ROZYYEV Turkmenistan 17.227 23.638 40.865 Bronze 117 Shaina KHAN Pakistan 20.955 19.891 40.847 Bronze 118 Petr JURIK Czech Republic 15.377 25.109 40.486 Bronze 119 Olha BALABON Ukraine 6.926 33.471 40.398 Bronze 120 Dan PENG Canada 14.376 25.672 40.048 Bronze 121 WILLIAM Indonesia 8.977 30.489 39.466 Bronze 122 Romāns ČAPLINSKIS Latvia 12.821 26.535 39.356 Bronze 123 Liudmila KUSHNIR Belarus 4.124 34.946 39.071 Bronze 124 Elise DUBOUE-DIJON France 13.223 25.577 38.800 Bronze 125 Batyr GARLYYEV Turkmenistan 11.647 27.108 38.755 Bronze 126 Chin Heng GAN Singapore 10.653 28.078 38.731 Bronze 127 Henrik SOENSTEBY Norway 19.194 19.266 38.460 Bronze 128 Immanuel Ilavarasan THOMAS India 7.641 30.614 38.255 Bronze Daniel Enrique Cardenas 129 Venezuela 17.315 20.688 38.003 Bronze ARMAS 130 Simone CALVELLO Italy 11.900 25.906 37.806 Bronze Thomas Ashton Christopher 131 New Zealand 15.165 22.590 37.755 Bronze WONG 132 Ivan OGIBALOV Estonia 20.033 17.698 37.731 Bronze 133 Khursand YOROV Tajikistan 6.462 31.250 37.712 Bronze 134 Ingrid Cristiana VREJA Romania 8.593 29.015 37.608 Bronze 135 Ivanka Rosenova ZHIVKOVA Bulgaria 12.109 25.325 37.435 Bronze 136 Taneli Toivo Hermanni RAJALA Finland 15.913 21.495 37.408 Bronze Muhamad Azri Muhamad 137 Malaysia 16.076 21.185 37.261 Bronze MARICAN 138 Gah Hung LEE Malaysia 13.741 23.358 37.098 Bronze Jorio Almino de Alencar Arrais 139 Brazil 10.298 26.788 37.086 Bronze MOTA 140 Michael James PLUNKETT New Zealand 9.372 27.179 36.551 Bronze 141 Lubica KRAUSKOVA Slovakia 15.294 21.194 36.488 Bronze 142 Nicholas MOULAF Australia 5.543 30.779 36.322 Bronze 143 Jaroslav ZAK United Kingdom 8.125 28.067 36.191 Bronze 144 Victor Lopez FERRANDO Spain 16.276 19.873 36.149 Bronze 145 Hubert KALAUS Austria 13.146 22.585 35.731 Bronze 146 Dogukan DIKMEN Turkey 4.730 30.941 35.672 Bronze 147 Irena MATKOVIČ Slovenia 6.191 29.430 35.621 Bronze 148 Kazuki YAMAGUCHI Japan 12.328 23.116 35.444 Bronze 149 Anita PABANI Pakistan 15.947 18.986 34.933 Bronze 150 Manuel Garcia RICARDO Cuba 10.006 24.846 34.852 Bronze 151 Max HAFLIGER Switzerland 5.936 28.640 34.575 Bronze

119 152 Sofia IZMAILOV United States 12.278 22.193 34.471 Bronze 153 Mathijs de JONG Netherlands 17.322 17.121 34.443 Bronze 154 Luciano Hector DI STEFANO Argentina 12.951 21.413 34.364 Bronze 155 Itamar Avraham David SHAMAI Israel 8.463 25.672 34.135 Bronze 156 Cristiana FANCIULLO Italy 15.728 17.914 33.643 Bronze 157 Muradov NURMUHAMMET Turkmenistan 7.700 25.691 33.391 Bronze 158 Pablo Gustavo LEVRAND Argentina 11.167 22.168 33.335 Bronze 159 Sergio Fonseca CHITICA Mexico 12.978 20.035 33.013 Honor. Mention 160 Roberta POCEVICIUTE Lithuania 8.430 24.572 33.002 Honor. Mention 161 Quentin LEFEBVRE France 4.698 28.219 32.917 Honor. Mention 162 Andreas FRUTIGER Switzerland 17.922 14.839 32.761 Honor. Mention 163 Manh Le DINH Vietnam 4.047 28.585 32.632 Honor. Mention 164 Amirhady KAMKARAMOLI Iran 9.323 23.252 32.575 Honor. Mention 165 Geir Haakon BECKSTROEM Norway 11.213 21.347 32.560 Honor. Mention 166 Javzansuren NORVANCHIG Mongolia 19.940 12.369 32.309 Honor. Mention 167 Umed BOLTAEV Tajikistan 6.581 25.623 32.204 Honor. Mention 168 Sabyrbek ZHEENTAEV Kyrgyzstan 11.125 21.073 32.198 Honor. Mention 169 Jose Enrique Robles SOTO Mexico 9.053 23.109 32.162 170 Vincenzo SPALLUTO Italy 19.643 12.481 32.124 171 Juraj AHEL Croatia 10.906 21.088 31.994 172 Dan Liraz LIDJI Israel 10.623 21.332 31.955 173 Christiaan Alwin DOUMA Netherlands 11.891 19.908 31.800 174 Alberto Garcia BOSQUE Spain 10.887 20.903 31.790 175 Luis Alberto Ypanaque ROCHA Peru 13.484 18.050 31.533 176 Andri Vilberg ORRASON Iceland 7.558 23.597 31.155 177 Jia CHEN Netherlands 13.756 17.073 30.830 178 Ivana BREKALO Croatia 6.885 23.812 30.698 179 Benjamin BOUSQUET France 10.071 20.369 30.440 180 Daria Ewa STRUSKA Sweden 13.226 17.060 30.285 181 Haleluya SAGIT Israel 14.383 15.827 30.209 182 Michalis ROSSIDES Cyprus 14.549 15.565 30.114 183 Martina FEYZRAKHMANOVA Ireland 6.592 23.499 30.091 184 Sotirios CHRISTODOULOU Greece 15.933 14.089 30.021 185 Mehmet VURAL Turkey 4.770 24.801 29.571 186 Aminatulmunirah KASIM Malaysia 14.217 14.921 29.138 Cristhian Luis Canari 187 Peru 10.133 18.935 29.069 CHUMPITAZ 188 Miha Emerik HABIC Slovenia 9.107 19.950 29.057 189 Ivan PREPOLEC Croatia 10.678 18.349 29.027 190 Andres Suarez VELAZQUEZ Spain 9.964 18.946 28.910 191 Zamirbek AKIMBEKOV Kyrgyzstan 6.850 21.749 28.599 192 Shervin GHAFOURI-TABRIZI Canada 5.318 22.533 27.852 193 Veaceslav VIERU Moldova 10.240 17.595 27.835 194 Stella Lucie RIAD Sweden 9.659 17.868 27.527 195 Urandelger TUVSHINDORJ Mongolia 15.717 11.531 27.248 Janne Valo Verner 196 Finland 14.805 11.925 26.729 HENRIKSSON 197 Konstantinos HADJIPETROU Cyprus 9.872 16.678 26.550 198 Patrick Michael O'SULLIVAN Ireland 7.094 18.803 25.897 199 Hordur Freyr YNGVASON Iceland 9.179 16.564 25.743 200 Pāvels ZUBOVIČS Latvia 7.373 17.985 25.358 201 Quentin HISETTE Belgium 15.442 9.809 25.251 202 Petr STADLBAUER Czech Republic 11.430 13.603 25.033 203 Karl NJALSSON Iceland 7.899 17.130 25.029 204 Ivan BARUN Croatia 5.171 19.810 24.981

120 205 Axel Nils Ola GOTTFRIES Sweden 10.125 14.801 24.927 206 John Christopher JANETZKO Canada 4.758 20.104 24.862 207 Johan Yuan WANG Norway 2.883 21.874 24.756 208 Tina KANSTRUP Denmark 8.823 15.645 24.468 209 Jose Ernando Sousa FILHO Brazil 2.863 21.546 24.409 210 Juan Ivan GOMEZ-PERALTA Mexico 9.277 14.958 24.234 211 Panagiotis PALANTAS Greece 5.106 18.826 23.932 212 Grigoris KATSIOLIDIS Cyprus 12.632 10.737 23.369 213 Charis Antonis ANTONIOU Cyprus 8.125 15.176 23.301 214 Ioannis BOTIS Greece 7.950 15.092 23.042 215 Grellan Jerome Kevin TUOHY Ireland 10.116 12.747 22.863 216 Vanessa LOODTS Belgium 6.211 16.556 22.767 217 Frederic COTTIER Switzerland 7.529 14.979 22.508 218 Alexandru CARTALEANU Moldova 5.053 17.351 22.403 219 Alberto Sanchez MOLERO Spain 9.554 12.339 21.893 220 Shahzad IQBAL Pakistan 4.009 17.060 21.069 Joseph Hernan Pena 221 Peru 4.056 16.917 20.973 ECHEVARRIA 222 Battulga BYAMBASUREN Mongolia 7.648 13.250 20.898 223 Leonid BIBIN Moldova 4.891 15.658 20.549 224 David Van CAUWENBERGE Belgium 6.442 14.073 20.515 225 Saidullo SULAYMONZODA Tajikistan 0.200 20.001 20.201 226 Hamza Khan SHAHBAZI Pakistan 3.236 16.374 19.61 227 Vesteinn SNAEBJARNARSON Iceland 8.010 11.492 19.502 228 Raul Joao de Sousa PEREIRA Portugal 1.700 17.775 19.475 229 Jānis JERMAKS Latvia 5.723 12.879 18.602 Carlos Leonel Ahumada 230 Mexico 4.001 14.163 18.164 MANUEL 231 Khusrav OLIMI Tajikistan 7.695 9.478 17.174 232 Christoffer NORN Denmark 6.830 9.317 16.147 233 Camilla ESPEDAL Norway 6.282 8.980 15.262 234 Stefan Dimitrov KADIYSKY Bulgaria 3.777 11.429 15.206 235 Linus Benjamin TÖRNQVIST Finland 5.883 8.921 14.803 236 Jorn WALSCHARTS Belgium 6.200 8.248 14.448 237 Mari Liisa TEINILÄ Finland 6.281 8.021 14.302 238 Gergana Sasheva VALCHEVA Bulgaria 0.930 13.367 14.297 239 Rui Emanuel Ferreira da SILVA Portugal 3.864 9.715 13.579 240 Joaquin GRASSI Uruguay 3.691 9.487 13.178 Rui Filipe Goncalves 241 Portugal 4.770 7.717 12.487 APOSTOLO 242 Memmed MIRZEYEV Azerbaijan 12.458 0.000 12.458 243 Argyroula LAVI Greece 7.598 4.483 12.081 244 Nicat MUSAYEV Azerbaijan 9.665 0.000 9.665 245 Sebastian BARCARDAL Uruguay 2.326 7.189 9.515 246 Rui Filipe Lebres LOPES Portugal 1.450 7.862 9.312 247 Uyanga DAGVADORJ Mongolia 2.093 6.712 8.805 248 Osamah Altaib SAFWAN Saudi Arabia 0.450 7.233 7.683 249 Diego OTERO Uruguay 3.388 4.266 7.655 250 Sebastian FIAMENE Uruguay 2.326 4.011 6.336 251 Mohammed Abdullah ALGASIM Saudi Arabia 1.960 4.323 6.283 252 Ibrahim Abdullah Ba JAAFR Saudi Arabia 1.928 3.844 5.772 253 Ataallah Naif ALHARBI Saudi Arabia 1.150 4.590 5.740 254 Abdulla AHMADOV Azerbaijan 5.365 0.000 5.365 255 Sohbet HOJAMUHAMMEDOV Turkmenistan 0.000 4.352 4.352 256 Orxan RZAYEV Azerbaijan 0.233 0.000 0.233

121 DETAILED RESULTS PER COUNTRY

Country Name P_1 P_2 T_1 T_2 T_3 T_4 T_5 T_6 T_7 T_8 Practice Theory Total Rank Medal Matias Daniel Gomez 4.963 10.483 5.571 3.636 4.846 5.024 3.200 2.838 3.103 2.105 15.446 30.324 45.771 85 Silver ELIAS Gonzalo Jose MUNAR 2.093 18.250 2.286 4.000 3.769 3.840 3.500 3.216 0.569 0.926 20.343 22.106 42.449 107 Bronze Argentina Luciano Hector 8.951 4.000 5.000 0.727 4.128 3.520 1.700 5.108 0.724 0.505 12.951 21.413 34.364 154 Bronze DI STEFANO Pablo Gustavo LEVRAND 6.167 5.000 3.857 6.182 3.410 5.120 0.000 0.946 0.000 2.653 11.167 22.168 33.335 158 Bronze Samvel BARDAKHCHYAN 13.605 1.400 6.714 7.091 7.000 3.456 6.200 2.270 7.500 1.642 15.005 41.874 56.878 34 Silver Armenia Narek DSHKHUNYN 1.395 2.950 7.000 8.000 7.000 5.888 6.700 4.541 7.500 2.147 4.345 48.776 53.121 48 Silver Arshavir 0.000 0.900 7.000 6.182 6.641 2.080 6.700 1.892 7.500 3.621 0.900 41.616 42.516 105 Bronze GHAHRAMANYAN Andrew TULLOCH 8.721 9.583 6.286 5.818 6.282 4.000 0.200 4.919 4.966 4.253 18.304 36.723 55.027 42 Silver Kartik RAMESH 7.255 5.510 5.857 3.455 4.128 5.280 2.800 4.919 4.397 3.958 12.765 34.793 47.558 71 Silver Australia Zijun (Jim) GE 2.209 5.267 4.143 4.909 4.487 5.280 2.100 4.919 3.362 4.674 7.476 33.874 41.350 114 Bronze Nicholas MOULAF 2.093 3.450 5.429 4.182 5.205 2.208 2.100 5.297 1.810 4.547 5.543 30.779 36.322 142 Bronze Alexander PUCHHAMMER 8.893 11.700 4.714 4.000 5.026 4.000 4.200 5.676 1.034 3.832 20.593 32.482 53.075 49 Silver Stephan PRIBITZER 19.111 6.017 3.571 3.273 4.128 3.040 1.700 4.162 1.810 0.800 25.128 22.485 47.613 70 Silver Austria Florian LANGMANN 4.419 7.450 5.286 2.000 3.410 2.880 3.600 4.730 6.766 2.989 11.869 31.661 43.529 97 Bronze Hubert KALAUS 6.746 6.400 3.286 0.364 4.128 1.920 1.700 1.514 7.190 2.484 13.146 22.585 35.731 145 Bronze Memmed MIRZEYEV 2.558 9.900 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 12.458 0.000 12.458 242 Nicat MUSAYEV 0.465 9.200 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 9.665 0.000 9.665 244 Azerbaijan Abdulla AHMADOV 0.465 4.900 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 5.365 0.000 5.365 254 Orxan RZAYEV 0.233 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.233 0.000 0.233 256 Dzianis HRAMAZDOU 8.605 13.900 5.857 5.636 4.308 6.880 3.700 4.162 0.414 2.821 22.505 33.778 56.283 37 Silver Alena VASKOVA 8.296 8.750 6.429 4.727 3.949 5.600 2.800 3.973 1.759 1.937 17.046 31.173 48.219 68 Silver Belarus Tsimafei BALOTNIK 11.447 3.900 4.000 3.273 5.205 4.160 3.500 3.500 0.569 1.811 15.347 26.017 41.364 113 Bronze Liudmila KUSHNIR 2.674 1.450 6.714 5.091 4.846 4.000 1.200 5.297 2.534 5.263 4.124 34.946 39.071 123 Bronze

122 Quentin HISETTE 5.642 9.800 4.143 0.727 0.359 1.600 0.700 1.324 0.155 0.800 15.442 9.809 25.251 201 Vanessa LOODTS 6.211 0.000 5.286 2.909 0.538 0.544 1.200 3.784 0.569 1.726 6.211 16.556 22.767 216 Belgium David Van 3.442 3.000 3.286 1.091 1.436 2.112 1.200 3.405 0.828 0.716 6.442 14.073 20.515 224 CAUWENBERGE Jorn WALSCHARTS 0.000 6.200 3.143 0.091 0.000 2.176 0.000 2.838 0.000 0.000 6.200 8.248 14.448 236 Thais Macedo Bezerra 3.605 3.200 5.000 5.091 5.205 4.160 5.000 5.297 7.138 3.368 6.805 40.260 47.064 75 Silver Terceiro JORGE Rafael de Cesaris Araujo 1.744 9.950 5.571 5.091 5.026 4.800 1.200 4.351 4.241 1.726 11.694 32.007 43.701 95 Bronze TAVARES Brazil Jorio Almino de Alencar 7.698 2.600 5.571 1.091 5.205 4.160 1.900 6.054 0.828 1.979 10.298 26.788 37.086 139 Bronze Arrais MOTA Jose Ernando Sousa 1.163 1.700 4.571 2.545 3.949 3.776 1.700 4.162 0.000 0.842 2.863 21.546 24.409 209 FILHO Zhivko Atanasov 9.158 17.350 1.429 2.364 4.487 4.608 2.700 4.635 0.310 3.158 26.508 23.691 50.199 58 Silver GEORGIEV Ivanka Rosenova 2.209 9.900 4.857 2.000 6.103 2.720 2.500 3.878 2.172 1.095 12.109 25.325 37.435 135 Bronze Bulgaria ZHIVKOVA Stefan Dimitrov KADIYSKY 2.977 0.800 2.000 0.364 3.410 1.920 2.500 0.000 0.983 0.253 3.777 11.429 15.206 234 Gergana Sasheva 0.930 0.000 5.286 1.818 0.718 2.464 1.000 2.081 0.000 0.000 0.930 13.367 14.297 238 VALCHEVA Hyeonjin BAE 12.307 14.333 5.286 5.273 4.487 1.504 3.500 4.351 2.793 4.168 26.640 31.362 58.003 29 Gold Dan PENG 10.476 3.900 5.143 4.545 5.744 1.574 1.700 3.595 0.466 2.905 14.376 25.672 40.048 120 Bronze Shervin GHAFOURI- Canada 4.208 1.110 6.714 2.182 5.923 3.200 1.000 1.703 1.138 0.674 5.318 22.533 27.852 192 TABRIZI John Christopher 3.558 1.200 4.714 0.182 4.487 0.320 2.700 2.459 3.052 2.189 4.758 20.104 24.862 206 JANETZKO Lei XU 3.721 16.050 7.000 8.000 7.000 7.200 5.900 6.811 7.190 7.200 19.771 56.300 76.071 1 Gold Yuan FANG 9.535 16.233 6.714 8.000 6.462 6.400 7.000 5.108 2.172 5.684 25.768 47.541 73.309 2 Gold China Zi-yang ZHANG 14.047 6.350 5.571 5.818 5.923 6.720 6.200 6.243 7.190 6.526 20.397 50.192 70.588 5 Gold Le YANG 4.279 9.450 6.714 8.000 6.641 5.728 5.100 7.000 6.921 3.032 13.729 49.136 62.865 14 Gold Ying Yu HO 6.485 16.900 6.714 7.636 6.103 5.440 5.700 4.919 6.672 3.832 23.385 47.016 70.401 6 Gold Chinese Wei-Lun HUANG 15.669 7.467 6.143 5.455 4.487 6.560 5.100 5.676 4.267 6.526 23.136 44.214 67.349 11 Gold Taipei Kai-Jui CHANG 10.925 6.267 5.286 5.818 3.949 5.280 2.900 5.865 5.431 5.474 17.192 40.002 57.194 32 Silver Frank Meng LIN 7.624 10.550 6.714 5.818 4.846 4.160 5.100 4.919 4.190 1.853 18.174 37.600 55.774 38 Silver

123 Juraj AHEL 3.256 7.650 5.571 2.455 2.692 1.120 1.200 3.973 0.414 3.663 10.906 21.088 31.994 171 Ivana BREKALO 2.385 4.500 6.143 1.455 4.487 3.360 2.800 3.784 0.310 1.474 6.885 23.812 30.698 178 Croatia Ivan PREPOLEC 6.894 3.783 4.143 1.091 4.308 6.240 2.000 0.568 0.000 0.000 10.678 18.349 29.027 189 Ivan BARUN 4.311 0.860 6.143 4.727 3.590 4.160 0.000 0.000 1.190 0.000 5.171 19.810 24.980 204 Cuba Manuel Garcia RICARDO 3.140 6.867 2.000 3.273 4.487 6.656 3.700 4.730 0.000 0.000 10.006 24.846 34.852 150 Bronze Michalis ROSSIDES 11.939 2.610 3.286 2.727 4.487 1.862 1.000 1.892 0.310 0.000 14.549 15.565 30.114 182 Konstantinos 8.372 1.500 4.429 3.636 2.872 3.200 1.300 0.568 0.000 0.674 9.872 16.678 26.550 197 Cyprus HADJIPETROU Grigoris KATSIOLIDIS 11.066 1.567 4.286 1.455 1.436 0.480 1.000 2.081 0.000 0.000 12.632 10.737 23.369 212 Charis Antonis ANTONIOU 2.558 5.567 4.143 0.818 2.333 2.528 2.200 3.027 0.000 0.126 8.125 15.176 23.301 213 Ctirad CERVINKA 17.386 10.300 5.857 6.182 3.410 2.880 1.000 3.027 1.345 3.747 27.686 27.448 55.134 41 Silver Czech Petr HOSEK 5.476 15.683 6.143 0.000 4.128 4.736 2.000 4.730 5.017 3.747 21.159 30.501 51.661 55 Silver Republic Petr JURIK 13.177 2.200 5.143 2.545 5.385 5.376 0.200 3.595 0.466 2.400 15.377 25.109 40.486 118 Bronze Petr STADLBAUER 3.930 7.500 1.429 0.000 2.333 2.560 2.900 2.838 0.828 0.716 11.430 13.603 25.033 202 Christian MARBOE 9.278 12.550 5.857 2.091 2.692 3.840 1.000 3.973 0.207 3.368 21.828 23.029 44.857 90 Bronze Thorbjørn Juul MORSING 10.659 13.500 4.286 0.727 4.846 2.560 1.000 3.784 0.414 0.337 24.159 17.954 42.112 111 Bronze Denmark Tina KANSTRUP 3.023 5.800 4.286 1.164 1.974 1.600 2.000 1.135 1.086 2.400 8.823 15.645 24.468 208 Christoffer NORN 6.830 0.000 1.571 0.727 0.718 2.886 1.900 1.514 0.000 0.000 6.830 9.317 16.147 232 Andres LAAN 9.955 8.800 6.143 2.364 5.744 4.800 5.200 3.973 6.103 3.242 18.755 37.569 56.324 36 Silver Andi KIPPER 12.059 10.500 4.714 0.909 5.026 4.480 0.500 3.784 5.017 1.095 22.559 25.525 48.084 69 Silver Estonia Taavi PUNGAS 3.907 10.650 5.429 5.455 5.205 1.760 2.200 4.730 0.414 2.905 14.557 28.097 42.654 104 Bronze Ivan OGIBALOV 11.733 8.300 4.429 1.091 4.308 1.120 0.700 3.784 1.552 0.716 20.033 17.698 37.731 132 Bronze Taneli Toivo Hermanni 6.047 9.867 5.571 1.818 3.949 3.136 2.400 2.649 0.414 1.558 15.913 21.495 37.408 136 Bronze RAJALA Janne Valo Verner 8.605 6.200 3.857 0.364 1.615 3.040 1.000 1.324 0.724 0.000 14.805 11.925 26.729 196 Finland HENRIKSSON Linus Benjamin 3.083 2.800 3.286 0.182 0.000 0.480 1.000 3.973 0.000 0.000 5.883 8.921 14.803 235 TÖRNQVIST Mari Liisa TEINILÄ 5.381 0.900 0.857 0.727 1.436 1.920 1.000 2.081 0.000 0.000 6.281 8.021 14.302 237 Simon GOURDIN-BERTIN 9.607 17.850 6.143 6.909 6.641 6.080 4.500 5.486 0.517 4.253 27.457 40.529 67.986 9 Gold Elise DUBOUE-DIJON 4.613 8.610 5.571 2.909 6.282 5.120 2.000 1.514 1.086 1.095 13.223 25.577 38.800 124 Bronze France Quentin LEFEBVRE 4.698 0.000 5.571 4.000 6.641 3.648 1.200 4.541 1.397 1.221 4.698 28.219 32.917 161 HM Benjamin BOUSQUET 8.721 1.350 3.857 1.818 4.846 1.702 2.000 3.595 1.034 1.516 10.071 20.369 30.440 179

124 Dimitry Ur'evic LOUTCHKO 13.796 12.317 5.000 7.818 5.564 4.768 7.000 4.919 6.983 2.063 26.113 44.115 70.228 7 Gold Philipp Albert STEININGER 17.707 13.750 5.857 4.000 5.564 5.280 1.000 3.595 1.241 4.084 31.457 30.621 62.078 16 Gold Germany Stefan Michael PUSCH 3.674 11.233 6.143 4.364 4.487 4.960 3.000 4.919 1.707 5.474 14.908 35.053 49.961 60 Silver Christian OBERENDER 7.851 7.100 4.571 4.545 4.128 6.720 1.800 4.351 0.931 4.421 14.951 31.469 46.420 81 Silver Sotirios CHRISTODOULOU 0.233 15.700 4.429 3.273 3.410 0.000 0.200 1.135 0.000 1.642 15.933 14.089 30.021 184 Panagiotis PALANTAS 1.163 3.943 5.286 2.000 1.974 0.320 5.500 1.135 0.000 2.611 5.106 18.826 23.932 211 Greece Ioannis BOTIS 0.000 7.950 5.429 0.364 5.923 0.000 0.200 1.703 0.000 1.474 7.950 15.092 23.042 214 Argyroula LAVI 0.698 6.900 0.857 1.091 0.359 0.870 0.000 0.000 0.000 1.305 7.598 4.483 12.080 243 Gyula PALFY 3.953 15.200 5.286 6.909 4.487 4.320 1.700 6.054 7.293 3.663 19.153 39.712 58.866 25 Gold Attila LOVAS 12.400 14.000 5.286 0.000 4.487 6.560 3.500 4.730 0.828 1.558 26.400 26.948 53.348 47 Silver Hungary Lorinc SARKANY 6.019 14.150 3.143 4.727 6.462 4.480 3.500 4.351 0.310 2.863 20.169 29.837 50.005 59 Silver Balint BALAZS 8.347 13.300 5.286 1.091 4.487 1.216 2.800 4.730 0.414 5.137 21.647 25.160 46.807 76 Silver Andri Vilberg ORRASON 2.558 5.000 6.286 4.000 4.846 3.936 0.000 2.649 0.828 1.053 7.558 23.597 31.155 176 Hordur Freyr YNGVASON 9.179 0.000 4.143 0.000 6.282 2.016 0.700 1.892 0.310 1.221 9.179 16.564 25.743 199 Iceland Karl NJALSSON 6.299 1.600 4.000 0.000 4.308 3.424 0.700 2.270 0.828 1.600 7.899 17.130 25.029 203 Vesteinn 0.000 8.010 1.714 1.091 2.872 2.560 0.500 1.703 0.000 1.053 8.010 11.492 19.502 227 SNAEBJARNARSON Sumit SOMANI 2.791 13.400 6.000 7.636 4.487 6.400 3.400 6.054 4.707 4.884 16.191 43.569 59.759 22 Gold Soham MEHTA 8.561 5.400 6.714 6.000 5.923 5.760 5.100 5.865 5.845 4.589 13.961 45.797 59.757 23 Gold India Vikas PRAJAPATI 1.814 12.000 6.429 8.000 4.846 6.560 3.200 6.243 2.690 2.063 13.814 40.031 53.845 46 Silver Immanuel Ilavarasan 5.891 1.750 6.429 7.455 6.282 1.280 1.900 4.351 0.517 2.400 7.641 30.614 38.255 128 Bronze THOMAS Teuku Mahfuzh Aufar KARI 2.728 16.700 5.429 5.818 5.205 3.200 1.700 2.932 3.310 2.779 19.428 30.374 49.802 61 Silver Muhamad FAIZ 10.228 2.000 6.714 5.636 5.923 3.680 2.100 5.486 1.655 5.347 12.228 36.543 48.770 66 Silver Indonesia Vincentius Jeremy 3.008 14.783 4.571 3.455 4.487 4.000 2.000 2.838 0.466 3.495 17.791 25.311 43.102 101 Bronze SUHARDI WILLIAM 6.927 2.050 4.857 6.545 5.923 4.704 1.300 3.973 1.966 1.221 8.977 30.489 39.466 121 Bronze Ehsan SHABANI 11.640 17.000 6.143 2.909 4.128 3.840 3.900 4.919 2.017 0.926 28.640 28.783 57.423 31 Gold Kaveh MATINKHOO 6.427 9.050 5.000 7.273 4.487 2.880 4.000 4.351 0.931 2.989 15.477 31.912 47.388 73 Silver Iran Shabnam SHARIFZADEH 8.605 6.850 5.286 4.364 4.128 3.040 4.500 4.351 0.517 1.979 15.455 28.165 43.620 96 Bronze Amirhady KAMKARAMOLI 3.023 6.300 5.429 3.091 3.410 3.680 0.000 3.216 0.931 3.495 9.323 23.252 32.575 164 HM

125 Andrew Brian CAIRNS 3.256 14.750 2.714 5.818 4.128 3.520 2.000 3.216 1.034 1.853 18.006 24.284 42.290 109 Bronze Martina 3.442 3.150 4.714 4.909 4.846 3.136 1.400 2.838 1.319 0.337 6.592 23.499 30.091 183 FEYZRAKHMANOVA Ireland Patrick Michael 1.860 5.233 4.857 3.273 4.487 2.720 1.200 1.703 0.310 0.253 7.094 18.803 25.897 198 O'SULLIVAN Grellan Jerome Kevin 3.256 6.860 3.857 0.182 2.692 0.736 1.000 1.703 0.724 1.853 10.116 12.747 22.863 215 TUOHY Assaf Avraham SHAPIRA 8.210 11.400 6.714 7.636 4.308 4.320 3.600 1.986 0.000 2.063 19.610 30.628 50.238 57 Silver Itamar Avraham David 2.163 6.300 5.143 6.545 5.923 4.096 1.000 2.459 0.000 0.505 8.463 25.672 34.135 155 Bronze Israel SHAMAI Dan Liraz LIDJI 8.140 2.483 3.857 2.182 6.282 0.480 2.500 3.405 2.121 0.505 10.623 21.332 31.955 172 Haleluya SAGIT 11.433 2.950 3.857 1.091 3.949 2.560 0.900 2.459 0.000 1.011 14.383 15.827 30.209 181 Vincenzo GRANDE 8.837 6.600 5.286 6.727 4.846 4.896 3.400 2.838 1.345 1.726 15.437 31.064 46.501 78 Silver Simone CALVELLO 4.650 7.250 3.571 3.636 4.487 3.200 2.700 5.297 1.034 1.979 11.900 25.906 37.806 130 Bronze Italy Cristiana FANCIULLO 11.195 4.533 6.429 0.000 3.410 3.840 0.000 2.838 0.724 0.674 15.728 17.914 33.643 156 Bronze Vincenzo SPALLUTO 7.093 12.550 1.143 1.091 2.333 3.040 1.600 0.946 2.328 0.000 19.643 12.481 32.124 170 Takashi HIROI 5.475 1.560 6.143 2.545 4.487 5.440 1.800 4.541 6.724 5.811 7.035 37.491 44.526 92 Bronze Naru TANAKA 4.853 1.600 6.429 4.182 6.282 2.720 4.500 5.486 2.679 4.632 6.453 36.910 43.363 100 Bronze Japan Shotaro TSUNODA 6.217 0.800 5.857 8.000 6.641 3.840 1.700 4.824 1.448 3.368 7.017 35.679 42.696 103 Bronze Kazuki YAMAGUCHI 9.778 2.550 6.571 2.182 3.769 1.536 2.500 3.595 0.310 2.653 12.328 23.116 35.444 148 Bronze Yerdos ORDABAYEV 13.721 7.200 6.143 5.818 7.000 4.256 2.100 3.973 0.724 1.558 20.921 31.572 52.493 52 Silver Ayana BADRAKOVA 4.975 11.900 5.000 7.091 5.923 4.320 0.000 3.405 0.000 1.558 16.875 27.297 44.172 93 Bronze Kazakhstan Dauren KALIYEV 10.066 8.367 1.714 1.091 2.333 6.336 4.100 3.973 2.534 1.516 18.433 23.598 42.031 112 Bronze Zhanbolat ZHOLGELDIYEV 14.302 0.450 4.357 2.182 4.308 2.880 0.800 4.257 5.121 2.568 14.752 26.473 41.225 115 Bronze Jae Soo KIM 19.000 16.000 5.857 5.636 6.641 3.520 1.800 5.486 0.517 5.263 35.000 34.721 69.721 8 Gold Korea Tae Gon OH 6.568 16.150 6.143 8.000 4.487 4.800 6.200 6.432 5.007 3.958 22.718 45.027 67.745 10 Gold Republic Chang Ho LEE 3.558 18.750 6.714 8.000 4.487 3.200 1.600 5.865 4.241 3.495 22.308 37.602 59.911 20 Gold Jong Soo YOON 3.755 6.410 6.429 6.364 5.923 2.400 5.000 4.351 5.897 2.737 10.165 39.100 49.265 64 Silver Kainar KAMALOV 7.501 5.543 5.571 3.818 3.231 4.736 5.500 2.649 6.517 1.011 13.044 33.033 46.077 82 Silver Kyrgyzstan Sabyrbek ZHEENTAEV 3.225 7.900 2.143 1.818 5.744 4.096 3.100 2.459 0.155 1.558 11.125 21.073 32.198 168 HM Zamirbek AKIMBEKOV 0.000 6.850 3.429 1.091 3.410 4.096 3.700 2.838 0.828 2.358 6.850 21.749 28.599 191

126 Mihails ARHANGELSKIS 5.463 1.600 5.714 5.091 4.308 4.640 5.400 5.676 4.086 2.989 7.063 37.904 44.967 89 Bronze Romāns ČAPLINSKIS 7.971 4.850 1.429 0.182 5.564 5.440 3.600 5.676 1.655 2.989 12.821 26.535 39.356 122 Bronze Latvia Pāvels ZUBOVIČS 3.023 4.350 5.286 3.455 5.923 2.464 0.500 0.189 0.000 0.168 7.373 17.985 25.358 200 Jānis JERMAKS 3.023 2.700 2.000 0.364 2.333 3.744 1.600 2.838 0.000 0.000 5.723 12.879 18.602 229 Karolis LEONAVICIUS 7.569 18.250 6.000 4.909 4.128 4.160 2.700 4.351 0.310 6.021 25.819 32.580 58.399 26 Gold Aurimas VYSNIAUSKAS 11.083 12.800 6.429 6.909 4.487 4.160 4.200 2.649 0.621 4.758 23.883 34.212 58.095 28 Gold Lithuania Bernardas MORKUNAS 6.730 5.817 6.000 6.545 3.769 3.936 3.200 4.351 3.828 5.432 12.547 37.061 49.608 63 Silver Roberta POCEVICIUTE 5.430 3.000 5.143 4.909 4.487 3.360 2.700 3.973 0.000 0.000 8.430 24.572 33.002 160 HM Jun Yan GOH 7.890 17.300 5.571 1.091 4.487 2.176 1.700 2.459 0.000 0.379 25.190 17.864 43.054 102 Bronze Muhamad Azri Muhamad 7.326 8.750 5.143 4.182 1.974 3.040 1.000 3.405 0.672 1.768 16.076 21.185 37.261 137 Bronze Malaysia MARICAN Gah Hung LEE 2.791 10.950 5.571 0.727 4.487 2.976 2.500 4.541 0.155 2.400 13.741 23.358 37.098 138 Bronze Aminatulmunirah KASIM 7.817 6.400 2.857 0.909 2.692 2.400 1.000 2.459 0.414 2.189 14.217 14.921 29.138 186 Sergio Fonseca CHITICA 1.628 11.350 3.571 1.091 3.949 4.800 2.100 3.689 0.414 0.421 12.978 20.035 33.013 159 HM Jose Enrique Robles SOTO 8.153 0.900 4.714 1.091 3.949 3.840 1.200 5.676 1.966 0.674 9.053 23.109 32.162 169 Juan Ivan GOMEZ- Mexico 1.977 7.300 1.714 0.727 4.846 3.200 1.900 1.703 0.362 0.505 9.277 14.958 24.234 210 PERALTA Carlos Leonel Ahumada 2.651 1.350 2.571 1.636 0.359 1.280 1.900 3.784 0.569 2.063 4.001 14.163 18.164 230 MANUEL Vasile GRAUR 8.030 18.610 5.000 4.000 3.769 4.640 1.600 2.081 0.000 2.063 26.640 23.153 49.793 62 Silver Veaceslav VIERU 8.140 2.100 3.429 0.364 1.795 4.960 1.800 3.973 0.517 0.758 10.240 17.595 27.835 193 Moldova Alexandru CARTALEANU 2.519 2.533 3.286 0.000 2.513 3.520 2.800 3.027 0.310 1.895 5.053 17.351 22.403 218 Leonid BIBIN 4.891 0.000 6.000 0.000 3.590 2.976 1.200 1.892 0.000 0.000 4.891 15.658 20.549 223 Javzansuren 13.023 6.917 3.857 3.273 0.538 2.560 0.500 1.135 0.000 0.505 19.940 12.369 32.309 166 HM NORVANCHIG Mongolia Urandelger TUVSHINDORJ 0.000 15.717 2.286 2.727 2.513 0.768 1.000 1.514 0.724 0.000 15.717 11.531 27.248 195 Battulga BYAMBASUREN 0.698 6.950 2.000 1.818 1.615 3.680 1.200 2.270 0.414 0.253 7.648 13.250 20.898 222 Uyanga DAGVADORJ 2.093 0.000 1.857 0.727 0.359 2.080 1.000 0.378 0.310 0.000 2.093 6.712 8.805 247 Raoul ROSENTHAL 7.773 16.800 1.429 2.909 4.667 2.240 2.200 3.595 0.776 1.137 24.573 18.952 43.525 98 Bronze Mathijs de JONG 4.372 12.950 2.143 1.818 3.590 4.160 1.500 3.784 0.000 0.126 17.322 17.121 34.443 153 Bronze Netherlands Christiaan Alwin DOUMA 8.558 3.333 5.143 0.727 3.231 4.288 1.500 2.459 1.086 1.474 11.891 19.908 31.800 173 Jia CHEN 7.173 6.583 4.000 0.182 2.692 2.112 1.700 2.459 0.517 3.411 13.756 17.073 30.830 177

127 Jonathan LIN 17.196 9.227 6.143 1.091 5.205 2.720 1.000 4.730 2.431 1.937 26.423 25.257 51.679 54 Silver Chuan Zheng LEE 5.221 16.000 6.143 0.182 3.769 3.520 1.200 4.919 0.000 6.147 21.221 25.880 47.101 74 Silver New Zealand Thomas Ashton Christopher 2.605 12.560 5.429 0.909 3.769 2.240 2.700 4.730 0.414 2.400 15.165 22.590 37.755 131 Bronze WONG Michael James PLUNKETT 8.372 1.000 5.429 5.818 4.128 4.384 2.000 3.027 0.414 1.979 9.372 27.179 36.551 140 Bronze Henrik SOENSTEBY 7.744 11.450 5.857 2.364 2.872 2.752 1.200 3.405 0.310 0.505 19.194 19.266 38.460 127 Bronze Geir Haakon 8.013 3.200 4.714 2.182 4.487 3.680 1.300 2.270 1.914 0.800 11.213 21.347 32.560 165 HM Norway BECKSTROEM Johan Yuan WANG 0.233 2.650 5.286 5.091 3.590 2.560 1.200 2.649 0.362 1.137 2.883 21.874 24.756 207 Camilla ESPEDAL 5.482 0.800 4.714 0.182 0.538 0.480 0.500 1.892 0.000 0.674 6.282 8.980 15.262 233 Shaina KHAN 8.755 12.200 2.000 3.273 1.974 3.680 1.500 2.081 1.552 3.832 20.955 19.891 40.847 117 Bronze Anita PABANI 8.937 7.010 4.000 0.909 3.410 4.800 1.500 2.649 1.086 0.632 15.947 18.986 34.933 149 Bronze Pakistan Shahzad IQBAL 3.209 0.800 4.571 1.091 2.513 3.328 1.000 2.838 0.414 1.305 4.009 17.060 21.069 220 Hamza Khan SHAHBAZI 1.186 2.050 5.429 1.091 2.692 1.120 1.000 3.216 0.310 1.516 3.236 16.374 19.610 226 Luis Alberto Ypanaque 13.484 0.000 4.429 0.545 0.359 2.240 0.000 3.027 6.776 0.674 13.484 18.050 31.533 175 ROCHA Cristhian Luis Canari Peru 0.000 10.133 1.714 0.364 6.282 3.776 1.600 3.595 0.931 0.674 10.133 18.935 29.069 187 CHUMPITAZ Joseph Hernan Pena 3.256 0.800 5.286 0.000 5.026 4.096 1.200 0.378 0.931 0.000 4.056 16.917 20.973 221 ECHEVARRIA Dawid Grzegorz LICHOSYT 19.256 18.550 4.571 1.091 4.846 3.872 4.400 4.730 1.759 2.189 37.806 27.458 65.264 12 Gold Krzysztof Cezary 9.609 18.600 6.429 1.818 2.333 4.000 3.700 5.865 4.955 6.400 28.209 35.500 63.710 13 Gold KOSINSKI Poland Wojciech Dominik MAGON 8.639 14.550 6.143 8.000 4.308 3.040 1.900 5.014 6.352 2.989 23.189 37.745 60.935 19 Gold Przemyslaw Krzysztof 10.667 8.910 5.571 8.000 4.128 6.240 2.300 4.730 0.931 6.105 19.577 38.006 57.582 30 Gold TREDAK Raul Joao de Sousa 0.000 1.700 4.429 1.818 3.410 3.616 1.200 2.459 0.000 0.842 1.700 17.775 19.475 228 PEREIRA Rui Emanuel Ferreira da 0.814 3.050 3.286 1.091 1.256 2.240 1.000 0.000 0.000 0.842 3.864 9.715 13.579 239 Portugal SILVA Rui Filipe Goncalves 4.170 0.600 0.571 1.091 0.897 2.656 0.700 1.135 0.414 0.253 4.770 7.717 12.487 241 APOSTOLO Rui Filipe Lebres LOPES 0.000 1.450 3.000 2.182 1.077 0.320 0.000 0.946 0.000 0.337 1.450 7.862 9.312 246

128 Ioan Teodor TROTUS 11.231 0.800 5.714 7.455 4.128 5.600 4.000 4.162 0.517 5.137 12.031 36.713 48.744 67 Silver Vlad Alexandru PUSCASU 9.149 7.050 5.857 4.727 6.282 3.616 4.100 5.297 1.448 0.000 16.199 31.328 47.527 72 Silver Romania Ionut Gabriel DUMITRU 6.315 4.820 6.571 2.909 3.949 3.040 6.000 3.595 6.259 3.032 11.135 35.354 46.489 79 Silver Ingrid Cristiana VREJA 2.093 6.500 6.286 3.091 5.744 2.080 3.500 4.919 0.828 2.568 8.593 29.015 37.608 134 Bronze Leonid ROMASHOV 19.256 13.300 6.143 7.273 3.949 3.526 5.600 5.108 3.776 4.337 32.556 39.712 72.267 3 Gold Russian Vasiliy VOROBYEV 13.340 14.860 6.000 4.000 3.769 7.360 3.700 5.865 6.103 6.611 28.200 43.408 71.609 4 Gold Federation Stanislav TEREHOV 13.464 11.750 5.429 5.455 4.846 4.800 3.100 7.000 3.103 3.832 25.214 37.564 62.778 15 Gold Eugeny NEKHOROSHEV 9.191 3.800 6.857 5.818 4.487 5.440 7.200 6.432 6.983 5.263 12.991 48.481 61.472 18 Gold Osamah Altaib SAFWAN 0.000 0.450 1.714 0.727 1.256 0.013 1.500 1.230 0.414 0.379 0.450 7.233 7.683 248 Mohammed Abdullah 0.000 1.960 2.000 0.000 0.359 0.160 0.500 1.135 0.000 0.168 1.960 4.323 6.283 251 Saudi Arabia ALGASIM Ibrahim Abdullah Ba JAAFR 1.628 0.300 0.286 0.727 0.359 0.006 1.000 0.757 0.414 0.295 1.928 3.844 5.772 252 Ataallah Naif ALHARBI 0.000 1.150 1.143 0.000 0.359 0.006 1.000 1.703 0.000 0.379 1.150 4.590 5.740 253 Zeqi YANG 12.015 4.383 6.143 0.182 4.487 4.960 4.800 4.730 5.586 5.305 16.398 36.193 52.591 51 Silver Benjamin CHEN 5.824 10.493 5.857 2.182 3.949 2.944 4.500 3.405 2.586 4.168 16.318 29.592 45.909 84 Silver Singapore Guang Jun Joseph LIM 2.326 9.100 5.857 1.818 4.308 3.840 3.300 4.351 4.707 2.737 11.426 30.918 42.344 108 Bronze Chin Heng GAN 2.093 8.560 5.857 1.818 3.410 0.960 3.800 4.162 3.776 4.295 10.653 28.078 38.731 126 Bronze Martin LUKACISIN 12.565 14.300 5.000 7.455 4.128 4.000 2.500 4.351 2.638 2.063 26.865 32.135 59.000 24 Gold Erik ANDRIS 12.030 16.177 5.571 1.455 4.846 5.792 3.800 3.784 2.741 0.505 28.207 28.495 56.701 35 Silver Slovakia Boris FACKOVEC 9.535 12.950 6.571 3.636 4.128 3.520 3.800 4.919 4.397 1.389 22.485 32.361 54.846 44 Silver Lubica KRAUSKOVA 3.561 11.733 5.000 2.364 4.487 3.296 1.000 3.405 0.000 1.642 15.294 21.194 36.488 141 Bronze Jan KOGOJ 9.488 2.467 6.143 8.000 4.487 4.576 1.000 4.351 2.534 3.663 11.955 34.755 46.710 77 Silver Jan BITENC 6.000 11.450 4.714 1.455 5.564 5.280 2.100 3.027 2.069 4.337 17.450 28.546 45.996 83 Silver Slovenia Irena MATKOVIČ 2.791 3.400 6.000 6.545 4.487 5.920 1.000 2.838 0.155 2.484 6.191 29.430 35.621 147 Bronze Miha Emerik HABIC 2.907 6.200 5.571 1.818 3.051 2.400 1.000 3.027 0.724 2.358 9.107 19.950 29.057 188 Victor Lopez FERRANDO 10.716 5.560 5.571 3.273 2.154 5.440 0.500 2.081 0.517 0.337 16.276 19.873 36.149 144 Bronze Alberto Garcia BOSQUE 2.837 8.050 5.286 1.818 1.436 3.840 1.000 5.486 0.310 1.726 10.887 20.903 31.790 174 Spain Andres Suarez 8.564 1.400 3.857 2.727 2.513 2.624 0.700 5.297 0.259 0.968 9.964 18.946 28.910 190 VELAZQUEZ Alberto Sanchez Molero 7.720 1.833 1.143 1.818 0.718 4.800 1.000 1.892 0.000 0.968 9.554 12.339 21.893 219

129 Oscar Carl Gunnar 8.140 15.350 5.571 3.818 5.744 0.480 1.000 3.878 0.517 1.179 23.490 22.188 45.677 86 Silver GRANBERG Sweden Daria Ewa STRUSKA 2.326 10.900 3.857 0.727 4.308 0.480 1.000 3.027 1.345 2.316 13.226 17.060 30.285 180 Stella Lucie RIAD 3.209 6.450 2.429 3.455 2.154 2.976 1.000 4.162 0.724 0.968 9.659 17.868 27.527 194 Axel Nils Ola GOTTFRIES 5.875 4.250 5.000 0.909 2.513 3.040 0.500 1.703 0.000 1.137 10.125 14.801 24.927 205 Lucia Domenica MEIER 7.442 15.200 3.571 3.091 3.590 3.200 0.700 3.595 0.000 1.895 22.642 19.641 42.283 110 Bronze Max HAFLIGER 5.936 0.000 6.143 2.545 4.667 1.984 1.500 3.405 5.069 3.326 5.936 28.640 34.575 151 Bronze Switzerland Andreas FRUTIGER 7.922 10.000 4.714 0.000 2.333 3.200 1.000 1.892 0.310 1.389 17.922 14.839 32.761 162 HM Frederic COTTIER 1.395 6.133 5.000 0.727 2.872 1.760 1.500 2.459 0.155 0.505 7.529 14.979 22.508 217 Khursand YOROV 1.395 5.067 4.286 6.909 5.385 6.720 2.200 4.824 0.000 0.926 6.462 31.250 37.712 133 Bronze Umed BOLTAEV 5.581 1.000 3.286 2.364 3.410 4.960 2.000 3.595 4.914 1.095 6.581 25.623 32.204 167 HM Tajikistan Saidullo SULAYMONZODA 0.000 0.200 4.000 2.818 4.487 2.368 1.200 4.351 0.776 0.000 0.200 20.001 20.201 225 Khusrav OLIMI 1.395 6.300 1.286 0.364 0.718 2.176 1.200 2.649 1.086 0.000 7.695 9.478 17.174 231 Bavorn HONGSRICHINDA 14.186 17.950 6.143 3.636 4.128 4.832 3.200 4.351 1.190 2.021 32.136 29.501 61.638 17 Gold Tanatorn 4.461 11.750 6.429 5.455 5.923 3.680 2.800 5.392 5.328 5.853 16.211 40.858 57.069 33 Silver Thailand KHOTAVIVATTANA Pornchai KAEWSAPSAK 4.186 10.400 6.714 4.545 5.923 3.526 6.000 6.149 0.517 2.484 14.586 35.859 50.445 56 Silver Techin CHULADESA 9.070 3.033 5.286 4.364 4.487 5.280 3.500 5.014 2.793 2.821 12.103 33.544 45.647 87 Silver Hande BOYACI 18.067 8.867 4.714 4.909 5.205 3.936 2.600 4.162 2.379 5.053 26.934 32.959 59.893 21 Gold Ahmet Selim HAN 9.767 4.633 4.857 5.273 4.487 3.680 4.900 0.757 2.534 2.484 14.401 28.972 43.373 99 Bronze Turkey Dogukan DIKMEN 0.930 3.800 5.143 3.091 5.205 4.416 2.500 4.541 3.983 2.063 4.730 30.941 35.672 146 Bronze Mehmet VURAL 1.860 2.910 5.571 2.182 1.974 2.560 3.500 3.973 3.103 1.937 4.770 24.801 29.571 185 Serdar ROZYYEV 1.977 15.250 5.000 1.727 1.795 5.600 2.900 2.838 2.431 1.347 17.227 23.638 40.865 116 Bronze Batyr Garlyyev 9.847 1.800 4.714 1.091 5.205 3.872 2.500 4.730 2.638 2.358 11.647 27.108 38.755 125 Bronze Turkmenistan Muradov NURMUHAMMET 0.000 7.700 4.286 3.636 5.564 3.040 0.500 4.730 1.914 2.021 7.700 25.691 33.391 157 Bronze Sohbet 0.000 0.000 1.143 1.091 0.000 0.640 1.100 0.378 0.000 0.000 0.000 4.352 4.352 255 HOJAMUHAMMEDOV Ostap CHERVAK 2.849 14.110 6.714 3.273 5.564 5.280 3.600 6.622 2.534 4.842 16.959 38.429 55.388 39 Silver Volodymyr TKACHENKO 2.698 10.333 5.286 0.364 5.564 5.280 2.500 3.595 5.948 2.316 13.031 30.852 43.883 94 Bronze Ukraine Kirill POLISCHUK 5.018 3.100 5.714 6.727 5.923 6.176 5.200 0.000 2.690 1.937 8.118 34.367 42.485 106 Bronze Olha BALABON 5.426 1.500 6.143 5.818 4.487 3.456 5.000 3.973 0.931 3.663 6.926 33.471 40.398 119 Bronze

130 Matthew James CLIFFE 9.070 5.860 5.286 6.182 5.205 4.672 3.300 6.811 1.552 5.137 14.930 38.144 53.074 50 Silver Frederick Robert William 7.105 10.793 6.857 8.000 4.128 4.480 1.200 3.027 5.586 1.263 17.898 34.542 52.440 53 Silver United Meath MANNERS Kingdom Derek Steven Hung-Che 4.885 5.900 6.143 7.273 5.205 5.760 1.900 4.730 0.517 2.484 10.785 34.012 44.797 91 Bronze CHAN Jaroslav ZAK 6.275 1.850 5.714 7.364 4.128 1.280 2.900 5.865 0.310 0.505 8.125 28.067 36.191 143 Bronze Brian Kihoon LEE 5.440 9.650 5.286 2.182 3.769 5.760 5.300 4.162 7.086 6.611 15.090 40.156 55.246 40 Silver Justin KOH 8.791 16.650 5.000 1.091 5.564 4.160 1.100 6.243 2.069 4.253 25.441 29.480 54.921 43 Silver United States Kenneth BREWER 2.151 12.500 5.571 7.273 6.641 3.680 1.600 5.108 0.000 4.337 14.651 34.210 48.861 65 Silver Sofia IZMAILOV 9.128 3.150 4.286 1.727 4.846 2.656 1.500 3.595 0.931 2.653 12.278 22.193 34.471 152 Bronze Joaquin GRASSI 2.791 0.900 2.143 1.818 0.538 0.320 1.200 2.743 0.724 0.000 3.691 9.487 13.178 240 Sebastian BARCARDAL 2.326 0.000 2.429 0.727 0.000 0.704 1.000 1.514 0.310 0.505 2.326 7.189 9.515 245 Uruguay Diego OTERO 2.488 0.900 0.286 0.364 0.718 1.280 0.200 1.419 0.000 0.000 3.388 4.266 7.655 249 Sebastian FIAMENE 2.326 0.000 0.286 0.727 0.000 1.120 1.000 0.568 0.310 0.000 2.326 4.011 6.336 250 Gabriel Eduardo Sanoja 11.734 15.100 3.143 3.273 3.769 2.880 1.000 2.838 1.448 0.000 26.834 18.351 45.185 88 Bronze LOPEZ Venezuela Daniel Enrique Cardenas 10.465 6.850 4.571 3.273 7.000 1.408 1.300 0.946 0.000 2.189 17.315 20.688 38.003 129 Bronze ARMAS Minh Nguyen Thi NGOC 2.698 8.500 5.857 7.091 7.000 7.040 5.000 5.676 7.190 2.189 11.198 47.043 58.241 27 Gold Ha Phan Tran HONG 8.605 3.400 5.857 7.818 6.282 5.440 3.500 5.770 6.672 1.095 12.005 42.435 54.439 45 Silver Vietnam Linh Bui LE 11.121 1.450 4.143 5.455 4.128 5.440 5.800 5.297 1.759 1.853 12.571 33.874 46.445 80 Silver Manh Le DINH 3.047 1.000 5.857 4.636 5.205 5.286 4.000 1.324 2.276 0.000 4.047 28.585 32.632 163 HM

131 Comparative Analysis of the IChO Results from 1990 to 2007

First First Last First Last First Last Δ(Bronze/Non- Year Country Δ(Gold/Silver) Δ(Silver/Bronze) Non- Gold Gold Silver Silver Bronze Bronze Medalists) Medalist 1990 France 82 65 2.75 62.25 57.25 1.25 56 42.5 1.5 41 1991 Poland 94.75 87.25 1.25 86 77 0.75 76.25 61.5 1.5 60 1992 USA 96.42 90.42 0.38 90.04 85.85 0.44 85.41 78.01 0.39 77.62 1993 Italy 83.5 75.5 2.5 73 64.5 0.5 64 49.5 1.5 48 1994 Norway 83.87 79.14 0.714 78.426 72.49 0.705 71.785 58.936 0.326 58.61 1995 China 95.541 82.115 0.32 81.795 73.027 0.322 72.705 60.124 0.002 60.122 1996 Russia 92.251 79.717 1.135 78.582 67.133 0.128 67.005 53.151 0.459 52.692 1997 Canada 81.25 72.5 0.5 72 65.75 0.75 65 54.5 0.5 54 1998 Australia 95.49 88.36 0.22 88.14 80.17 0.65 79.52 65.6 0.86 64.74 1999 Thailand 94.118 81.609 1.249 80.36 70.061 0.145 69.916 53.158 0.203 52.955 2000 Denmark 94.46 80.313 0.468 79.845 71.914 0.341 71.573 58.201 0.5 57.701 2001 India 92.31 82.91 1.1 81.81 68.61 0.27 68.34 50.13 0.73 49.4 2002 Netherlands 92.5 81.35 2.23 79.12 72.14 0.3 71.84 61.45 0.51 60.94 2003 Greece 96.43 87.83 0.3 87.53 80.07 1.04 79.03 65.13 0.99 64.14 2004 Germany 88.7 72.1 1.1 71 60 0.4 59.6 47 0.9 46.1 2005 Taiwan 96.75 90.6 0.89 89.71 80.62 0.54 80.08 64.07 0.85 63.22 2006 Korea 93.43 70.61 0.48 70.13 56.71 0.32 56.39 39.41 0.17 39.24 2007 Russia 76.71 57.423 0.229 57.194 45.647 0.462 45.185 33.335 0.322 33.013

The 39th IChO Organizers acknowledge Dr. Wolfgang Hampe for the idea of the statistical analysis and contribution of the data

132 List of Mentors, Observers, and Guests

Country Role Name Jorge Alberto Onofrio HM Bruno Argentina M Maria Laura Uhrig HM Lida Sahakyan Armenia M Artak Tovmasyan HM Anthony Edward Phillips Australia M Alex Chit Hei Wong SO Mark John Ellison HM Manfred Kerschbaumer Austria M Lisbeth Berner HM Vaqif Abbasov M Mutellim Abbasov Azerbaijan SO Yusif Abdullayev SO Nasim Ajdar Abishov HM Viktar Khvaluk Belarus M Aliaksandr Rahoisha HM Sebastien Delfosse M Hans L.S. Vanhoe Belgium Cedric Pascal Patrick SO Malherbe HM Sergio Maia Melo Jose Arimateia Dantas M Brazil Lopes Lucia Souza Carvalho SO Melo

133 HM Donka Nikolova Tasheva Bulgaria M Penka Vasileva Tsanova HM Andrew Paul Dicks Canada M Stanislaw Skonieczny HM Lianyun Duan M Ke-qing Zhao China SO Ying-xia Wang SO Bi-qin Wang

G Tai-Shan Fang G I-Hsing Chen Chinese Taipei HM I-Jy Chang M Bih-Yaw Jin SO Tun-Cheng Chien SO Ya-Ling Chen HM Branka Zorc Croatia M Tomislav Cvitas Luis Enrique Guerra Cas- HM Cuba tano HM Anaxagoras Hadjiosif Cyprus M Stella Ioannou-Loucaides Czech Repub- HM Eva Muchova lic M Petr Slavicek HM Kurt Bjoenager Nielsen Denmark M Hanne Busk SO Morten Foverskov HM Uno Maeorg Estonia M Vladislav Ivanishtshev HM Jorma Kullervo Koskimies M Markku Rafael Sundberg Finland Nina Helmi Katariina SO Aremo

134 HM Adrien Sebastien Meglio France M Guillaume Meriguet SO Vincent Tejedor HM Sabine Ingeborg Nick Germany M Wolfgang Hampe, StD SO Carsten Schmuck HM Anastasia Detsi Greece M Dimitrios Chiniadis HM Gyorgy Tarczay M Szilard Varga Hungary SO Laszlo Turi SO Attila Villanyi SO Andras Kotschy Gisli Holmar Johannes- HM son Iceland M Sigurdur Vidir Smarason SO Margret Lilja Bjornsdottir HM Dilip Kumar Maity Sambasivan Venkat Es- M India waran Swapna Mahesh SO Narvekar

HM Riwandi Sihombing Indonesia M Djulia Onggo SO Ismunaryo Moenandar HM Mansour Abedini M Ebrahim Kianmehr Iran Seyed Ali Seyedi Esfa- SO hani SO Mahin Jabalameli

135 HM Paraic James Ireland M Wesley Richard Browne

HM Moris S. Eisen Israel M Iris Barzilai HM Mario Anastasia Italy M Pietro Allevi HM Yoshiyuki Sugahara M Noriyuki Yonezawa Japan SO Keijiro Taga SO Masatada Matsuoka HM Kurmangali Bekishev Kazakhstan M Rassima Sadakbayeva G Eui seo Park G Duckhwan Lee G Sookyon, Yeo Lee Korea Republic HM Jung Hag Park M Tai Jong Kang SO Seonghoon Lee SO Hee Gweon Woo M Barak Mehdi Hadi Kuwait M Fotouh Alshamali HM Minira Batkibekova Kyrgyzstan M Raina Asakeeva HM Ināra Akmene Latvia M Skaidrite Pakule HM Rimantas Raudonis Lithuania M Edvinas Orentas

136 G Mohd Shah Noriah G Turiman Punia Malaysia HM Mohd Jamil Maah M Noorsaadah A. Rahman SO Mei Leng Lee Carlos Mauricio Castro- HM Acuna Mexico Eugenio Octavio Reyes M Salas Nadejda Gheorghe HM Moldova Velisco M Andrei Mihail Bunescu HM Dorj Daichaa Mongolia M Nyamgerel Choijilsuren SO Davaasuren Sandag HM Peter de Groot Netherlands M Emiel de Kleijn SO Cornelis Beers Suzanne Margaret Boni- HM New Zealand face M David Salter G Olayemi Albert Bamikole Nigeria G Grace Oni Ojo SO Sunday Asher Adedeji HM Hans-Petter Hersleth Norway M Kristian Vestli HM Khalid Mohammed Khan Pakistan M Muhammad Shaiq Ali

137 HM Bertha Beatriz Flores Alor Peru M Galina Shevtsova HM Marek Orlik Poland M Janusz Stepinski Diana Claudia Gouveia HM Alves Pinto Portugal Maria do Amparo Ferreira M Faustino SO Alzira Pinto Rebelo

HM Marius Andruh Romania M Mihaela Maria Hillebrand SO Daniela Elisabeta Bogdan HM Sergey Igorevich Kargov Russian Fed- M Alexandr Belov eration SO Alexey Zeifman Mohammed Abdulkarim G Ibrahim Ghanem Saad Al- HM Saudi Arabia Ghanem M Hadi Ali Bahari SO Hassan Ahmed Masslouf HM Karen K.W. Mak Singapore M Basheer Chanbasha SO Loy Chuan Chua HM Anton Sirota Slovakia M Marta Salisova SO Jan Reguli HM Darko Dolenc Slovenia M Andrej Godec

138 Juan Antonio Rodriguez HM Renuncio Spain Maria Carmen Cartagena M Causape SO Fernando Latre David HM Per Henning Lindgren M Ulf Charles Jaglid Sweden SO Anna Cecilia Stenberg HM Maurice Cosandey Switzerland M Thomas Engeloch HM Abdufatokh Hotamov Tajikistan Golibsho Takdirovich M Nasymov G Narongsil Thooppanom HM Vudhichai Parasuk Thailand M Yongsak Sritana-Anant SO Amarawan Intasiri SO Ekasith Somsook HM Jale Hacaloglu Turkey M Cihangir Tanyeli HM Guvanchmyrat Paytakov Turkmenistan M Aman Begliyev HM Galyna Malchenko Ukraine M Yuriy Kholin

HM Charles Stuart McCaw United M Timothy Graham Hersey Kingdom SO Henneli Greyling SO William Peter Nolan

139 G Katherine Ringland Kotz G Ronald Edward Monson United States HM John Carl Kotz M Kara Anne Pezzi SO John Leon Kiappes Uruguay HM William Stebniki HM Amalia Torrealba Venezuela M Eliseo Silva G Hung Pham Tuan G Nga Nguyen Thi Vietnam HM Hue Tran thanh M Hien Pham dinh SO Dau Nguyen Van

140 IChO-2007 BUDGET US $ 1. Total Budget of the IChO-2007 2 956 200 1.1. Governmental source 1 813 600 1.2. Sponsors 1 070 000 1.3. Participation fees 72 600

2. Expenditures of the IChO-2007 2 956 200 2.1. Examination preparation 1 111 700 2.1.1. Equipment 595 500 2.1.2. Reagents 203 200 Practical and theoretical prepara- 2.1.3. 65 000 tory and examination tasks Facilities for the International Jury 2.1.4. 248 000 meetings and translation 2.2. Accomodation and Food 860 000 2.2.1. Students 380 000 2.2.2. Mentors 480 000 2.3. Transportation 154 700 2.3.1. Students 98 000 2.3.2. Mentors 56 700 2.4. Opening and Closing Ceremonies 120 000 2.5. Cultural program 68 000 2.5.1. Students 44 000 2.5.2. Mentors 24 000 2.6. Secretariat and operation 178 600 2.6.1. Staff costs 162 800 2.6.2. Equipment and services 15 800 2.7. Guides 42 000 2.8. Public relations 207 200 2.8.1. Catalyzer 60 000 2.8.2. Souvenirs 81 200 2.8.3. Presentations, mass media 66 000 2.9. The 39th IChO web-server 34 100 2.10. Final report 16 000 2.11. Consumables 12 300 2.12. Other 151 600

141 Participation fees for the IChO-2007

No. Country Amount, US $ 1 Argentina 1300 2 Armenia 200 3 Australia 800 4 Austria 2000 5 Azerbaijan 800 6 Belarus 1200 7 Belgium 2000 8 Brazil 900 9 Bulgaria 2000 10 Canada 1000 11 China 1200 12 Chinese Taipei 200 13 Croatia 800 14 Cuba 1600 15 Cyprus 1800 16 Czech Republic 1500 17 Denmark 700 18 Estonia 1400 19 Finland 1900 20 France 1700 21 Germany 300 22 Greece 400 23 Hungary 2000 24 Iceland 600 25 India 600 26 Indonesia 800 27 Iran 1500 28 Ireland 1100 29 Israel 200 30 Italy 1400 31 Japan 500 32 Kazakhstan 1000 33 Korea 100 34 Kuwait 1500

142 35 Kyrgyzstan 800 36 Latvia 1700 37 Lithuania 1700 38 Malaysia 200 39 Mexico 1600 40 Moldova 100 41 Mongolia 200 42 Netherlands 500 43 New Zealand 1600 44 Nigeria 45 Norway 1300 46 Pakistan 200 47 Peru 400 48 Poland 1600 49 Portugal 500 50 Romania 2000 51 Russia 0 52 Saudi Arabia 200 53 Singapore 1800 54 Slovakia 1500 55 Slovenia 1700 56 Spain 1200 57 Sweden 2000 58 Switzerland 2000 59 Tajikistan 400 60 Thailand 800 61 Turkey 1400 62 Turkmenistan 600 63 U.S.A. 1500 64 Ukraine 1400 65 United Kingdom 2000 66 Uruguay 67 Venezuela 1500 68 Vietnam 1200 Total Amount 72600

143 Preparations

2005 January– Choose the Venue: Chemistry Department, M.V. Lomonosov Moscow State University February – Choose the Period of the IChO December – Participation in the International Steering Committee Meeting, Korea

2006 February – Science Committee started its work July, 4 – Order of Prime-Minister of Russian Federation July – visit to the 38th IChO in Yeungnam University, Korea September – Order of Ministry of Education and Science, Russian Federation October – Order of Rector of M.V. Lomonosov Moscow State University October, 20 – Internet-site launched December, 7-10 – International Steering Committee Meeting, Moscow December – IChO 2007 Tentative Program

2007 January – Official Invitation January – Budget Funding from State Budget January-April – Laboratory Renovation February, 1 – Upload Preparatory Problems on Internet February – On-Line Registration Launched February-June – On-Line Registration February – Select Guides and Staff July, 15 to 24 – IChO December – Report to the International Steering Committee, Budapest

144 People who made the IChO-2007 possible

Organizing Committee 1 Petr Anisimov (Federal Agency for Education, RF) Grigorii Balykhin (Federal Agency for Education, RF) Tatiana Beshenenko (Federal Agency for Education, RF) Nikolai Bulaev (State Duma, RF) Evgenii Butko (Federal Agency for Education, RF) Svetlana Demidova (Federal Agency for Education, RF) Vadim Eremin Elena Eremina Andrei Fursenko – Chairman, Minister of Education and Science, RF Alexander Gladilin Isaak Kalina (Ministry of Education and Science, RF) Lyudmila Kokhanova Nikolai Kuz’menko Valerii Lunin – President of the IChO-2007 Vladimir Mironov (Vice-Rector, MSU) Elena Pazyuk Victor Shtepa Lyubov’ Strel’nikova (Chief Editor, Khimiya I Zhizn’) Vladimir Terenin

Science Committee

Ivan Babkin Anna Bacheva Yurii Barbalat Mikhail Beklemishev Svetlana Bendrysheva Anna Berkovitch Zhanna Boeva Alexander Bogachev Andrei Cheprakov Vadim Eremin, Co-Chair Bulat Garifullin (Bashkirian Medical Andrei Garmash State University) Alexander Gladilin, Co-Chair Il’ya Glebov Alexander Kisin (Institute of Chemistry and Eugene Karpushkin Technology of Organoelement Compounds)

1 Affiliation by default: Chemistry Department of MSU

145 Mikhail Korobov Alexey Korovin Sergei Legotskii Nikolay Melik-Nubarov Kirill Oskolok Nina Pasekova Valery Putlyaev Marina Reshetova Igor Sedov (Kazan’ State University, Marina Rozova A.Butlerov Institute of Chemistry) Irina Seregina Sergey Seryakov Mikhail Statkus Boris Tarasevich Igor Trushkov Igor Tyulkov Julia Valeeva Andrei Vedernikov (Univ. of Maryland)

Secretariat

Anna Bacheva Anna Berkovich Anastasiya Chekanova Elena Eremina Elena Pazyuk Boris Pokrovskii (web) Mikhail Tabunov (web) Ekaterina Yakubovich

Technical Committee

M. Belyakov A. Bibin M. Chudakova A. Demidova A. Ezhov M. Galkin An. Gladilin D. Ivanov K. Ivanova P. Kebets A. Marinchuk A. Natikan M. Nikitina A. Poteryaev E. Rodina M. Steklov V. Valaeva S. Vatsadze A. Veresov

Lab instructors

N. Aryutkina P. Binevskiy P.Chelushkin A. Chemagin M. Chernobrovkin G. Dudina O. Fedyanina V. Figurovskaya E. Ivanainen S. Ivanova S. Kurzeev M. Livantsov L. Livantsova I. Men’shikova I. Naperova O. Popova V. Sergeeva I. Shender O. Starostina T. Suslenkova G. Ushakov S. Vatsadze Yu. Volkova I. Zorov

146 Lab assistants

G. Demchuk A. Druzhinina D. Dzhigailo T. Klimashina N. Krutova S. Kubyshev P. Kudan L. Lukanina O. Lukanina T. Lyskova S. Matusova T. Matyushina E. Mishukova O. Monogarova N. Potapova L. Sergeeva L. Shadskaya L. Shimko M. Tereshina K. Yablotskiy A. Zatirakha

Spectrophotometer operators

O. Anikina O. Bogomolova V. Bugrin D. Davydov A. Filatov L. Filatova E. Gerasimova A. Gopin N. Gruzinskaya A. Kislukhin M. Kondrashov N. Kovaleva D. Maltseva M. Nemykh D. Pavlov I. Protasov E. Sbrueva T. Semashko V. Skorkin Yu. Smirnova M. Smolov A. Sokolyuk V. Spiridonov S. Uglanova V. Vasil’yev A. Zhirnov

“Catalyzer” team

A. Agaeva E. Bredina K. Davydova M. Eremina A. Galieva M. Kokhanova A. Kravtsova A. Lobus (photo) A. Lukichev (head) I. Lukichev M. Nefedieva A. Pomerantseva I. Salynkina S.Sobolev V. Turin

Opening and Closing Ceremonies designed by Victor Shiryaev

Doctors – Olga Eremina and Tatiana Pogorelova

147 Guides

Argentina Anna Kovrigina Armenia David Ayrapetyan Australia Nadezhda Kolesnik Austria Daniil Zaonegin Azerbaijan Eldar Rizaev Belarus Vitaliy Sushkevich Belgium Alexey Rozov Brazil Kate Abramova Bulgaria Georgi Valeriev Stoychev Canada Artem Kovalenko China P.R. Jin Zhao Chinese Taipei Anna Lyamina Croatia Anastasiya Galanina Cuba Anna Kalyushnaya Cyprus Elena Ivanova Czech Republic Kate Zaharevich Denmark Nadezhda Ogneva Estonia Eugenia Tamyar Finland Alexey Shikhantsov France Etienne Richard Germany Zhenya Kazanova Greece Anton Tunin Hungary Alexei Kulagin Iceland Svetlana Khoronenkova India Sergey Kovalevskiy Indonesia Marina Frolova Iran Tatiana Kononova Ireland Maxim Abakumov Israel Ksenia Sarycheva Italy Marina Ulanovskaya Japan Maria Khrenova Kazakhstan Timur Zhiyentayev Korea Republic Natalia Leshakova Kyrgyzstan Artem Skabeev Latvia Maxim Homutov Lithuania Dasha Yovcheva

148 Malaysia Alexandra Malishkina Mexico Nina Kozhemyakina Moldova Andrey Istrate Mongolia Ganbaatar Tsetserleg Netherlands Elena Galysheva New Zealand Kate Yakubovich Norway Natalia Morukova Pakistan Alexander Kuznetsov Peru Lidia Bogatyreva Poland Evgeniya Zhukovskaya Portugal Valentina Pyatickh Romania Alexey Godina Russian Federation Oleg Chernov Saudi Arabia Samir Zheltisov Singapore Mila Shtepa Slovakia Yana Kresan Slovenia Maria Kuleshova Spain Olga Usovich Sweden Kate Zakharevich Switzerland Daniil Troshinkin Tajikistan Ulmas Zhumaev Thailand Diana Kfuri Turkey Kyiyalbek Kaparov Turkmenistan Anton Reshetnyak Ukraine Maxim Zabilskiy United Kingdom Mikhail Sheybe United States Daria Tsareva Uruguay Azalia Korunbaeva Venezuela Anastasiya Nevokshanova Vietnam Nguen Din Tyyong mentors Anna Dyachenko mentors Alexandr Veresov mentors Igor Tyulkov mentors Oleg Brilev guests Aleksandra Prokhorova

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